Method of Integration
In this chapter, we first describe how we can reverse the differentiation. Then we explain the following:
10.1 Antiderivatives
Consider
f ' (x) = 3x2.
The natural question aries, what is the function f (x) ? Namely, what is f in terms of x? We know that differentiation decreases
the power by 1, so f must contain x3.
If f (x) = x3, then f ' (x) = 3x2, or
If f (x) = x3 + 2, then f ' (x) = 3x2, or
If f (x) = x3 - 1⁄2,
then f ' (x) = 3x2, etc.
It means that there are many such functions of the form
f (x) = x3 + C
where C is an arbitrary constant.
We say that
x3 is the antiderivative of 3x2.
Next consider
f ' (x) = x.
We think the same way as above, the original function f must contain x2. However,
d⁄dx = x2 = 2x,
we see the extra factor 2. If we multiply both sides by
1⁄2, then
d⁄dx (
1⁄2 x2) = x.
If f (x) = 1⁄2 x2 + C where C is an arbitrary
constant, then f ' (x) = x, so 1⁄2 x2 is the
antiderivative of x.
If F (x) is a function where F ' (x) = f (x), then the antiderivative of f (x) is
F (x).
We call the antiderivative as integral and write
∫ x dx = 1⁄2 x2 + C,
where C is the constan of integration.
We read this as “the integral of x with respect to x.”
In general,
if F ' (x) = f (x) then ∫ f (x) dx = F (x) + C
where dx means that the integration is taking place with respect to the variable x.
Here f (x) is called the integrand. The variable of integration in an integral plays no essential role. It might be x, or t, or u, or anything else:
∫ f (x) dx, ∫ f (t) dt, ∫ f (u) du, etc.
Example 1.
Find the antiderivative of
(a) √ x
(b) 3x5
(c) e3x
(d) 1⁄x3.
Solution
Since d⁄dx x3⁄2 = 3⁄2 x1⁄2 and d⁄dx ( 2⁄3 x3⁄2 ) = x1⁄2 = √ x ,
the antiderivative of √ x is 2⁄3 x3⁄2 .
(b) Since d⁄dx x6 = 6x5 and
d⁄dx ( 1⁄2 x6 ) = 3x5,
the antiderivative of 3x5 is 1⁄2 x6.
(c) Since d⁄dx e3x = 3e3x and
d⁄dx ( 1⁄3 e3x ) = e3x ,
the antiderivative of e3x is 1⁄3 e3x.
(d) Since d⁄dx x-2 = -2x-3 and
d⁄dx (- 1⁄2x2 ) = 1⁄x3 ,
the antiderivative of 1⁄x3 is - 1⁄2x2 .
We now describe the integration of fundamental functions. We know that
d⁄dx xn + 1 = (n + 1)xn .
The reverse of this process is
∫ xn dx = 1⁄n + 1 xn + 1 + C
for n ≠ -1.
When n = 0, it is seen that
∫ x0 dx = ∫ 1 dx = ∫ dx = x + C.
∫ dx = x + C.
When n = -1, let us consider the differentiation of ln x.
Since d⁄dx ln x = 1⁄x, it follows that
∫ 1⁄x dx = ln x + C, for x > 0.
For x < 0,
d⁄dx ln |x| = d⁄dx ln (-x) = 1⁄-x (-1)
= 1⁄x , by Chain Rule.
∫ 1⁄x dx = ln |x| + C, x ≠ 0
Rules of Integration
Suppose f (x) and g (x) are continuous functions and k ∈ ℝ
1. ∫ k dx = kx + C.
2. ∫ k f (x) dx = k ∫ f (x) dx.
3. ∫ [ f (x) ± g (x) ] dx = ∫ f (x) dx ± ∫ g (x) dx.
Example 2.
Evaluate each of the following integrals.
(a) ∫ (4x5 + 1) dx
(b) ∫ (2x6 - 1⁄3 x3 + 3⁄x) dx
(c) ∫ 5x √ x dx
(d) ∫ (x - 1)2⁄ √ x dx.
Solution
(a) ∫ (4x5 + 1) dx
= 4 ∫ x5 dx + ∫ 1 dx
= (4⁄6 x6 + C1) + (x + C2); (C1 and C2 are constants of integration)
= (2⁄3 x6 + x) + C1 + C2
= (2⁄3 x6 + x) + C. (C = C1 + C2 is another constant of integration)
From now on, we shall write constant of integration only in the answer.
(b) ∫ (2x6 - 1⁄3 x3 + 3⁄x ) dx
= 2 ∫ x6 dx - 1⁄3 ∫ x3 dx + 3 ∫ 1⁄x dx
= 2⁄7 x7 - 1⁄12 x4 + 3 ln |x| + C.
(c) ∫ 5x √ x dx
= 5 ∫ x x1⁄2 dx
= 5 ∫ x3⁄2 dx
= 5 (2⁄5 x5⁄2 ) + C.
= 2x5⁄2 + C.
(d) ∫ (x - 1)2⁄ √ x dx = ∫ x2 - 2x + 1⁄2 x1⁄2 dx
= ∫ (x2⁄ x1⁄2 - 2x⁄x 1⁄2 + 1⁄ x 1⁄2 ) dx
= ∫ x3⁄2 dx - 2 ∫ x 1⁄2 dx + ∫ x- 1⁄2 dx.
= 2⁄5 x5⁄2 - 4⁄3 x3⁄2 + 2x1⁄2 + C.
Integrating Exponential Function
d⁄dx ax = ax ln a, ∫ ax dx = 1⁄ln a ax + C, where a > 0, a ≠ 1.
When a = e,
d⁄dx ex = ex, ∫ ex dx = ex + C.
When a = e,
d⁄dx ex = ex, ∫ ex dx = ex + C.
Example 3.
Find the following integrals.
(a) ∫ e-x + 1⁄e-x dx
(b) ∫ (ex + 2x) dx
(c) ∫ (ex log 5 + 1⁄x2 ) dx
(d) ∫ 3x ln 3 dx.
Solution
(a) ∫ e-x + 1⁄e-x dx = ∫ (1 + ex) dx
= ∫ 1 dx + ∫ ex dx = x + ex + C.
(b) ∫ (ex + 2x) dx = ∫ ex dx + ∫ 2x dx
= ex + x2 + C.
(c) ∫ (ex log 5 + 1⁄x2 ) dx = ∫ ex log 5 dx + ∫ x-2 dx
= ex log 5 - 1⁄x + C.
(d) ∫ 3x ln 3 dx = 1⁄ln 3 3x Ln 3 + C
= 3x + C.
Integrating Trigonometric Functions
d⁄dx sin x = cos x,
∫ cos x dx = sin x + C.
d⁄dx (-cos x) = -(-sin x) = sin x, ∫ sin x dx = -cos x + C.
d⁄dx tan x = sec2 x, ∫ sec2 x dx = tan x + C.
d⁄dx (-cos x) = -(-sin x) = sin x, ∫ sin x dx = -cos x + C.
d⁄dx tan x = sec2 x, ∫ sec2 x dx = tan x + C.
Example 4.
Evaluate.
(a) ∫ (3 cos x - 5 sin x) dx
(b) ∫ (ex + 2 sin x) dx
(c) ∫ ex - √ x ⁄2 dx
(d) ∫ (1 + tan2 x) dx
Solution
(a) ∫ (3 cos x - 5 sin x) dx = 3 ∫ cos x dx - 5 ∫ sin x dx
= 3 sin x + 5 cos x = C.
(b) ∫ (ex + 2 sin x) dx = ∫ ex dx + 2 ∫ sin x dx
= ex - 2 cos x + C.
(c) ∫ ex - √ x ⁄2 dx = 1⁄2 ∫ ex dx - 1⁄2 ∫ x1⁄2 dx
= 1⁄2 ex - 1⁄3 x3⁄2 + C.
(d) ∫ (1 + tan2 x) dx = ∫ sec2 x dx
= tan x + C.
Integrating f (ax + b)
We have learned the reverse process of differentiation as an integration. Now we consider the integral of function which used the chain rule for differentiation.Consider the integral
∫ (4x + 1)5 dx.
By the chain rule,
d⁄dx (4x + 1)6 = 6(4x + 1)5 d⁄dx (4x + 1)
= 24 (4x + 1)5.
Multiplying by 1⁄24 on both sides, we get
1⁄24 d⁄dx (4x + 1)6 = (4x + 1)5.
d⁄dx ( 1⁄24 (4x + 1)6 ) = (4x + 1)5.
Therefore
∫ (4x + 1)5 dx = 1⁄24 (4x + 1)6 + C.
For n ≠ -1,
d⁄dx ( 1⁄a(n + 1) (ax + b)n + 1 ) = (ax + b)n.
∫ (ax + b)n dx = 1⁄a
(ax + b)n + 1 ⁄(n + 1) + C, n ≠ -1.
When n = -1,
d⁄dx ( 1⁄a ln (ax + b)) = 1⁄a ( a⁄ax + b )
= 1⁄ax + b , a ≠ 0.
∫ 1⁄ax + b dx =
1⁄a ln |ax + b| + C.
In general, if f is a differentiable functiion of x, then
d⁄dx f (ax + b) = f ' (ax + b) d⁄dx (ax + b)
= af ' (ax + b)
and reversing this we get
∫ f ' (ax + b) dx = 1⁄a f(ax + b) + C.
For the trigonometric functions of the form f (ax + b),
∫ cos(ax + b) dx = 1⁄a sin (ax + b) + C,
∫ sin (ax + b) dx = - 1⁄a cos (ax + b) + C,
∫ sec2 (ax + b) dx = 1⁄a tan (ax + b) + C.
∫ sin (ax + b) dx = - 1⁄a cos (ax + b) + C,
∫ sec2 (ax + b) dx = 1⁄a tan (ax + b) + C.
For the exponential functions of the form f (px + q),
∫ apx + q dx = 1⁄p
1⁄ln a apx + q + C
where a > 0, a ≠ 1.
When a = e,
∫ epx + q dx = 1⁄p epx + q + C.
When a = e,
∫ epx + q dx = 1⁄p epx + q + C.
Example 5.
Evaluate the following integrals.
(a) ∫ 1⁄3 - 4x dx
(b) ∫ √ 8x - 7 dx
(c) ∫ e2x + 1 dx
(d) ∫ 23x + 1 dx
(e) ∫ cos2 x dx
(f) ∫ sin 4x cos 3x dx
Solution
(a) ∫ 1⁄3 - 4x dx = 1⁄4 |3 - 4x| + C.
(b) ∫ √ 8x - 7 dx = 1⁄8 [ 2⁄3 (8x - 7)1⁄2 ] + C
= 1⁄12 (8x - 7)1⁄ 2 + C.
(c) ∫ e2x + 1 dx = 1⁄2 e2x + 1 + C.
(d) ∫ 23x + 1 dx = 1⁄3 1⁄ln 2 23x + 1 + C.
(e) We use cos2 x= 1⁄2 (1 + cos 2x), then
∫ cos2 x dx = 1⁄2 ∫ (1 + cos 2x) dx
= 1⁄2 (x + 1⁄2 sin 2x) + C
= 1⁄2 x + 1⁄4 sin 2x + C.
(f) We use sin 4x cos 3x = 1⁄2 (sin 7x + sin x), then
∫ sin 4x cos 3x dx = 1⁄2 ∫ (sin 7x + sin x) dx
= 1⁄2 (-1⁄7 cos 7x - cos x) + C
= - 1⁄14 cos 7x - 1⁄2 cos x + C.
Exercise 10.1
- Evaluate the following integrals:
- Evaluate the following integrals:
(a) ∫ 4x8 dx
(b) ∫ 3 ⁄ 2 x2 ∛ x dx
(c) ∫ (5x + 2) dx
(d) ∫ sin2 x dx
(e) ∫ x + 3 ⁄ √ x dx
(f) ∫ ( 1 ⁄ 2x + 5) dx
(g) ∫ ( ex + 2 ⁄ x ) dx
(h) ∫ ( 1 ⁄ x5 + 4ex ) dx
(i) ∫ ( 3 ⁄ x + ex + 10 )dx
(j) ∫ sin2 3x dx
(k) ∫ sin 5x sin 2x dx
(l) ∫ cos 7x cos 4x dx
(a) ∫ (1 - 2x)3 dx
(b) ∫ sin (2πx + 7) dx
(c) ∫ cos (3x - 7) dx
(d) ∫ 35x - 2 dx
(e) ∫ 1 ⁄ 7x - 6 dx
(f) ∫ sin 2x ⁄ sin x dx
(g) ∫ sec2(2x + 3) dx
(h) ∫ e7x - 3 dx
(i) ∫ (1 + tan2 2πx) dx
10.2 Substitution Method
We use the substitution method when the given integral can be transformed to the simpler integral by a change of variable. We describe how to get the integrals
$$\int f(g(x))g'(x) dx$$ and
$$\int \frac{g'(x)}{g(x)} dx$$
where g is differentiable and f is continuous. Then, we apply this method to the integral containing the trigonometric functions.
Let y = f(x) be a differentiable function. The differential dx is an independent variable, the differential dy is given by
dy = f' (x) (dx).
Here, dy is a dependent variable that depends on x and dx.
$$1. \int f(g(x))g'(x) dx $$
Suppose the function g is differentiable and f is continuous.
Let u = g(x). Then du = g' (x) dx, then
$$\int f(g(x))g'(x) dx = \int f(u) du.$$
To compute the integral by substitution method, the basic steps are as follows:
Step 1 Select a substitution u = g(x).
Step 2 Differentiate the substitution and arrange to write dx in term of du.
Step 3 Substitute the expression from step 2, and transform the entire
integral from x-variable to u-variable form.
Step 4 Integrate with respect to u.
Step 5 Rewrite the answer in terms of x.
Step 1 Select a substitution u = g(x).
Step 2 Differentiate the substitution and arrange to write dx in term of du.
Step 3 Substitute the expression from step 2, and transform the entire
integral from x-variable to u-variable form.
Step 4 Integrate with respect to u.
Step 5 Rewrite the answer in terms of x.
Example 6.
Evaluate each of the following integrals.
(a) ∫ xex2 dx
(b) ∫ sin4 x cos x dx
(c) ∫ x2 ( x3⁄3 + 1 )3 dx
(d) ∫ √ 1 - 5x dx
(e) ∫ 3x √ x2 + 5 dx
(f) ∫ cos3 x dx
Solution
(a) ∫ xex2 dx
Let u = x2. Then du = 2x dx,
1⁄2 du = x dx.
∫ xex2 dx = ∫ ex2 x dx
= ∫ eu 1⁄2 du
= 1⁄2 eu + C
= 1⁄2 ex2 + C.
(b) ∫ sin4 x cos x dx .
Let u = sin x. Then du = cos x dx .
∫ sin x4 cos x dx = ∫ u4 du
= u5⁄5 + C = sin5 x⁄5 + C.
(c) ∫ x2 ( x3⁄3 + 1 )3 dx
Let u = x3⁄3 + 1. Then du = x2 dx.
∫ x2 ( x3⁄3 + 1 ) 3 dx = ∫ u3 du
= u4⁄4 + C
= 1⁄4 ( x3⁄3 + 1 )4 + C.
(d) ∫ √ 1 - 5x dx
Let u = 1 - 5x. Then du = -5 dx,
- 1⁄5 du = dx.
∫ √ 1 - 5x dx = ∫(1 - 5x)1⁄2 dx
= ∫ u1⁄2 (- 1⁄5) du
= - 1⁄5 ⋅ 2⁄3 u3⁄2 + C
= - 2⁄15 u3⁄2 + C
= - 2⁄15(1 - 5x)3⁄ 2 + C.
(e) ∫ 3x √ x2 + 5 dx
Let u = x2 + 5. Then 1⁄2du = x dx
∫ 3x √ x2 + 5 dx = 3 ∫ (x2 + 5) 1⁄2 x dx
= 3 ∫ u1⁄2 1⁄2 du
= 3⁄2 ⋅ 2⁄3 u3⁄2 + C
= u 3⁄2 + C
= (x2 + 5)3⁄2 + C.
(f) ∫ cos3 x dx
We use cos2 x = 1 - sin2 x, then
∫ cos3 x dx = ∫ (1 - sin2 x) cos x dx.
Let u = sin x, then du = cos x dx
∫ cos3 x dx = ∫ (1 - u2) du
= u - u3⁄3 + C
= sin x - 1⁄3 sin3 x + c.
2. ∫ g ' (x)⁄g (x) dx
Let u = g (x). Then du = g ' (x) dx.
Thus
∫ g ' (x)⁄g (x) dx = ∫ 1⁄u du = ln |u| + C.
∫ g ' (x)⁄g (x) dx = ln |g (x)| +
C.
∫ tan x dx = ∫ sin x⁄cos x dx = - ln |cos x| + C.
∫ cot x dx = ∫ cos x⁄sin x dx = ln |sin x| + C.
∫ tan x dx = ∫ sin x⁄cos x dx = - ln |cos x| + C.
∫ cot x dx = ∫ cos x⁄sin x dx = ln |sin x| + C.
Example 7.
Evaluate the following integrals.
(a) ∫ 2x - 1⁄x2 - x - 6 dx
(b) ∫ x2 + 2x - 1⁄x2 - 1 dx
(c) ∫ 1⁄1 + ex dx
Solution
(a) ∫ 2x - 1⁄x2 - x - 6 dx
Let u = x2 - x - 6. Then du = (2x - 1) dx.
∫ 2x - 1⁄x2 - x - 6 dx = ∫ 1⁄u du
= ln |u| + C = ln |x2 - x - 6| + C.
(b) ∫ x2 + 2x - 1⁄x2 - 1 dx
We can rewrite as
∫ x2 + 2x - 1⁄x2 - 1 dx = ∫ (1 + 2x⁄x2 - 1 ) dx
Let u = x2 - 1. Then du = 2x dx.
$$\int \frac{x^2 + 2x -1}{x^2 - 1} dx = \int \left ( 1 + \frac{2x}{x^2 - 1} \right ) dx $$ = ∫ dx + ∫ 1⁄u du
= x + ln |u| + C
= x + ln |x2 - 1| + C.
(c) ∫ 1⁄1 + ex dx
We can rewrite by multiplying and dividing by e-x.
∫ 1⁄1 + ex dx = ∫ 1⁄1 + ex e-x⁄e-x dx
= ∫ e-x⁄e-x + 1 dx.
Let u = e-x + 1. Then du = -e-x dx.
∫ 1⁄1 + ex dx = ∫ e-x⁄e-x + 1 dx
= - ∫ 1⁄u du
= - ln |u| + C.
= - ln (e-x + 1) + C.
We now explain some integrals involving the trigonometric functions using method of substitution.
(a) ∫ sec x dx = ln | sec x + tan x + C
Proof
We have
∫ sec x dx = ∫ sec x sec x + tan x⁄sec x + tan x dx
= ∫ sec2 x + sec x tan x⁄sec x + tan x dx.
Let u = sec x + tan x. Then
du = (sec x tan x + sec2 x) dx.
So,
∫ sec x dx = ∫ 1⁄u du
= ln |u| + C
= ln |sec x + tan x| + C.
(b) ∫ csc x dx = - ln |csc x + cot x| + C
Proof
We have
∫ csc x dx = ∫ csc x csc x + cot x⁄csc x + cot x dx
= ∫ csc2 x + cot x csc x⁄csc x + cot x dx.
Let u = csc x + cot x. Then
du = - (cot x csc x + csc2 x) dx.
So,
∫ csc x dx = - ∫ 1⁄u du = - ln |u| + C
= - ln |csc x + cot x| + C.
Exercise 10.2
- Integrate the following functions using the given substitutions.
- Use the substitution method to evaluate the following integrals.
- Evaluate the integral ∫ x ⁄ (x2 + 1) ln (x2 + 1) dx.
(a) 4x3 √ x4 - 1 ; u = x4 - 1
(b) cos3 x sin x ; u = cos x
(c) 1 ⁄x ln |x| ; u = ln |x|
(d) sin5 x cos x; u = sin x
(e) ln x ⁄ x , x > 0; u = ln x
(f) x3ex4 ; u = x4
(a) ∫ x √ 1 - x dx
(b) ∫ (2x + 1)(x2 + x)7 dx
(c) ∫ sin3x dx
(d) ∫ x2 √ x3 - 2 dx
(e) ∫ sec2 x ⁄ tan x dx
(f) ∫ x ⁄ √ x + 1 dx
10.3 Integration by Parts
We use the method of integration by parts to integrate the product of two functions. In Section 10.1, we explain that the integral of the sum of functions is
the sum of respective integrals. But the integral of the product function is not the product of respective integrals. Therefore, we use another technique
and it is called the integration by parts.
It based on product rule of differentiation,
(u v)' dx = u ' v + u v'
where u and v are functions of x. Then integration both sides, we get
∫ (u v)' dx = ∫ u' v dx + ∫ u v' dx.
Applying the antiderivative in left hand side, we get
u v = ∫ u' v dx + ∫ u v' dx
or
∫ u v' dx = u v - ∫ u' v dx.
Since u ' dx = du and v' dx = dv, we get
∫ u dv = u v - ∫ v du.
To compute the integral of product of two functions using integration
by parts, the basic steps are as follows:
Step 1 Choose u and dv.
Step 2 Differentiate u and integrated dv.
Step 3 Substitute the expression from step 2 in
∫ u dv = uv - ∫ v du.
Step 4 Simplify.
by parts, the basic steps are as follows:
Step 1 Choose u and dv.
Step 2 Differentiate u and integrated dv.
Step 3 Substitute the expression from step 2 in
∫ u dv = uv - ∫ v du.
Step 4 Simplify.
Example 8.
Evaluate each of the following integrals.
(a) ∫ xex dx
(b) ∫ ln x dx
(c) ∫ x cos x dx
(d) ∫ x sin x dx
(e) ∫ ex sin x dx
(f) ∫ x2 ln x dx
Solution
(a) ∫ xex dx
Let u = x, dv = ex dx.
Then du = dx, v = ∫ ex dx = ex.
∫ xex dx = xex - ∫ exdx
= xex - ex + C.
(b) ∫ ln x dx
Let u = ln x, dv = dx.
Then du = 1⁄x dx , v = ∫ dx = x.
∫ ln x dx = x ln x - ∫ x 1⁄x dx + C
= x ln x - x + C.
(c) ∫ x cos x dx
Let u = x, dv = cos x dx.
Then du = dx, v = ∫ cos x dx = sin x.
∫ x cos x dx = x sin x- ∫ sin x dx
= x sin x + cos x + C.
(d) ∫ x sin x dx
Let u = x, dv = sin x dx.
Then du = dx,
v = ∫ sin x dx = - cos x.
∫ x sin x dx = -x cos x - ∫ (-cos x) dx
= -x cos x + sin x + C.
(e) ∫ ex sin x dx
Let u = ex, dv = sin x dx.
Then du = ex dx, v = ∫ sin x dx = -cos x.
∫ ex sin x dx = -ex cos x - ∫ (-cos
= -ex cos x + ∫ ex cos x dx.
Let I = ∫ ex sin x dx. Then we have
I = -ex cos x + ∫ ex cos x dx (1)
Let u = ex, dv = cos x dx.
Then du= ex dx, v = ∫ cos x dx = sin x.
Thus
∫ ex cos x dx = ex sin x - ∫ ex sin x dx. (2)
Substituting (2) into (1), we get
I = -ex cos x + ex sin x - ∫ ex sin x dx
= ex (sin x - cos x) - I
I = 1⁄2 ex (sin x - cos x) + C.
Therefore
∫ ex sin x dx = 1⁄2 ex (sin x - cos x) + C.
(f) ∫ x2 ln x dx
Let u = ln x, dv = x2 dx
Then du = 1⁄x dx,
v = ∫ x2 dx = x3⁄3.
∫x2 ln x dx = x3⁄3 ln x - ∫ x3⁄3 1⁄x dx
= x3⁄3 ln x - 1⁄3 x3⁄3 + C.
= x3 ln x⁄3 - x3⁄9 + C.
Exercise 10.3
- Use the integration by parts to evaluate the following integrals.
(a) ∫ s e-2s ds
(b) ∫ ln(x + 1)dx
(c) ∫ t sin 2t dt
(d) ∫ x 2x dx
(e) ∫ x cos 5x dx
(f) ∫ ex cos x dx
10.4 Partial Fraction Method
We use the partial fraction method to integrate the rational functions.
Example 9.
Evaluate the integral
∫
1⁄x2 + 2x - 3 dx.
Solution
$$\int \frac{1}{x^2 + 2x -3} dx$$ We rewrite the rational fraction as
$$\frac{1}{x^2 + 2x - 3} = \frac{1}{(x + 3)(x - 3)}.$$ However, we want this fraction as the sum of linear frractions as below.
$$\frac{1}{x^2 + 2x - 3} = \frac{A}{x + 3} + \frac{B}{x - 1} $$ where A and B are constants that will be determined later. Then, it can be rewritten as
$$\frac{1}{x^2 + 2x - 3} = \frac{A(x-1) + B(x+3)}{(x + 3)(x - 3)}.$$ Therefore,
$$1 = A(x-1) + B(x+3)$$ $$1 = (A + B)x + (-A + 3B).$$ Equating the coefficients of corresponding powers of x, we get
$$A + B = 0$$ $$-A + 3B = 1$$ which give $$A = -\frac{1}{4}$$ and $$B = \frac{1}{4}.$$ Therefore $$\frac{1}{x^2 + 2x - 3} = \frac{-\frac{1}{4}}{x + 3} + \frac{\frac{1}{4}}{x - 1} $$ Thus $$\int \frac{1}{x^2 + 2x - 3} dx =-\frac{1}{4} \frac{1}{x + 3} dx + \frac{1}{4} \int \frac{1}{x - 1} dx $$ $$= -\frac{1}{4} ln |x+3| + \frac{1}{4} ln |x-1| + C$$ $$= \frac{1}{4} ln \left | \frac{x-1}{x+3} \right | + C.$$
$$\int \frac{1}{x^2 + 2x -3} dx$$ We rewrite the rational fraction as
$$\frac{1}{x^2 + 2x - 3} = \frac{1}{(x + 3)(x - 3)}.$$ However, we want this fraction as the sum of linear frractions as below.
$$\frac{1}{x^2 + 2x - 3} = \frac{A}{x + 3} + \frac{B}{x - 1} $$ where A and B are constants that will be determined later. Then, it can be rewritten as
$$\frac{1}{x^2 + 2x - 3} = \frac{A(x-1) + B(x+3)}{(x + 3)(x - 3)}.$$ Therefore,
$$1 = A(x-1) + B(x+3)$$ $$1 = (A + B)x + (-A + 3B).$$ Equating the coefficients of corresponding powers of x, we get
$$A + B = 0$$ $$-A + 3B = 1$$ which give $$A = -\frac{1}{4}$$ and $$B = \frac{1}{4}.$$ Therefore $$\frac{1}{x^2 + 2x - 3} = \frac{-\frac{1}{4}}{x + 3} + \frac{\frac{1}{4}}{x - 1} $$ Thus $$\int \frac{1}{x^2 + 2x - 3} dx =-\frac{1}{4} \frac{1}{x + 3} dx + \frac{1}{4} \int \frac{1}{x - 1} dx $$ $$= -\frac{1}{4} ln |x+3| + \frac{1}{4} ln |x-1| + C$$ $$= \frac{1}{4} ln \left | \frac{x-1}{x+3} \right | + C.$$
Example 10
$$\text{Evaluate \, the \, integral \,} \int \frac{1}{2x^2 + 3x + 1} dx.$$
Solution
$$ \int \frac{1}{2x^2 + 3x + 1} dx.$$ First we write $$ \frac{1}{2x^2 + 3x + 1} = \frac{1}{(2x + 1)(x + 1)} = \frac{A}{2x + 1} + \frac{B}{x + 1}$$ and so $$\frac{1}{2x^2 + 3x + 1} = \frac{A(x + 1) + (B(2x + 1)}{(2x + 1)(x + 1)}$$ Therefore, $$1 = A(x + 1) + B(2x + 1).$$ when x = 1,
$$1 = B(-1), \text{so}\, \, B = -1.$$ $$\text{When\,} x = -\frac{1}{2}, \, \, 1 = A(\frac{1}{2}), \text{so}\, \, A = 2.$$ $$\int \frac{1}{2x^2 + 3x + 1} dx = \int \frac{2}{2x + 1} dx + \int \frac{-1}{x + 1} dx $$ $$ = ln |2x + 1| - ln |x + 1| + C$$ $$= ln \left | \frac{2x + 1}{x + 1} \right | + C.$$
$$ \int \frac{1}{2x^2 + 3x + 1} dx.$$ First we write $$ \frac{1}{2x^2 + 3x + 1} = \frac{1}{(2x + 1)(x + 1)} = \frac{A}{2x + 1} + \frac{B}{x + 1}$$ and so $$\frac{1}{2x^2 + 3x + 1} = \frac{A(x + 1) + (B(2x + 1)}{(2x + 1)(x + 1)}$$ Therefore, $$1 = A(x + 1) + B(2x + 1).$$ when x = 1,
$$1 = B(-1), \text{so}\, \, B = -1.$$ $$\text{When\,} x = -\frac{1}{2}, \, \, 1 = A(\frac{1}{2}), \text{so}\, \, A = 2.$$ $$\int \frac{1}{2x^2 + 3x + 1} dx = \int \frac{2}{2x + 1} dx + \int \frac{-1}{x + 1} dx $$ $$ = ln |2x + 1| - ln |x + 1| + C$$ $$= ln \left | \frac{2x + 1}{x + 1} \right | + C.$$
Example 11.
$$\text{Evaluate\,the\,integral\,}\, \int \frac{x^3 + x^2 - 4x}{x^2 -4} \,dx.$$
Solution
$$ \int \frac{x^3 + x^2 - 4x}{x^2 -4} \,dx$$ We can rewrite as $$ \int \frac{x^3 + x^2 - 4x}{x^2 -4} \,dx = \int \left( x+1 + \frac{4}{x^2 - 4}\right)\,dx.$$ First, we write $$\frac{4}{x^2 - 4} = \frac{4}{(x+2)(x-2)} = \frac{A}{x+2}+\frac{B}{x-2}$$ and so $$\frac{4}{x^2-4} = \frac{A(x-2) + B(x+2)}{(x+2)(x-2)}.$$ Therefore, $$4= A(x-2) + B(x+2).$$ $$\text{When\,} x = 2,\, \, \, 4=B(4), \, \, \, \text{so}\, \, B=1.$$ $$\text{When\,} x = -2,\, \, \, 4=A(-4), \, \, \, \text{so}\, \, A=-1.$$ $$\int \frac{x^3 + x^2 -4x}{x^2-4}\,dx = \int (x+1)\,dx + \int \frac{-1}{x+2}\,dx + \int \frac{1}{x-2}\,dx$$ $$= \frac{x^2}{2} + x-ln|x+2| + ln|x-2| + C$$ $$= \frac{x^2}{2} + x + ln \left|\frac{x-2}{x+2}\right| + C.$$
$$ \int \frac{x^3 + x^2 - 4x}{x^2 -4} \,dx$$ We can rewrite as $$ \int \frac{x^3 + x^2 - 4x}{x^2 -4} \,dx = \int \left( x+1 + \frac{4}{x^2 - 4}\right)\,dx.$$ First, we write $$\frac{4}{x^2 - 4} = \frac{4}{(x+2)(x-2)} = \frac{A}{x+2}+\frac{B}{x-2}$$ and so $$\frac{4}{x^2-4} = \frac{A(x-2) + B(x+2)}{(x+2)(x-2)}.$$ Therefore, $$4= A(x-2) + B(x+2).$$ $$\text{When\,} x = 2,\, \, \, 4=B(4), \, \, \, \text{so}\, \, B=1.$$ $$\text{When\,} x = -2,\, \, \, 4=A(-4), \, \, \, \text{so}\, \, A=-1.$$ $$\int \frac{x^3 + x^2 -4x}{x^2-4}\,dx = \int (x+1)\,dx + \int \frac{-1}{x+2}\,dx + \int \frac{1}{x-2}\,dx$$ $$= \frac{x^2}{2} + x-ln|x+2| + ln|x-2| + C$$ $$= \frac{x^2}{2} + x + ln \left|\frac{x-2}{x+2}\right| + C.$$
Example 12
$$\text{Evaluate\,the\,integral}\, \int \frac{2x+1}{(x-1)^2} \,dx.$$
Solution
$$\int \frac{2x+1}{(x-1)^2} \,dx.$$ We can rewrite as $$ \frac{2x+1}{(x-1)^2} = \frac{A}{x-1} + \frac{B}{(x-1)^2}$$ and so $$ \frac{2x+1}{(x-1)^2} = \frac{A(x-1)+B}{(x-1)^2}$$ Therefore, $$2x+1 = A(x-1) + B,$$ $$= Ax + (-A+B).$$ Equating the coefficients of corresponding powers of x, we get $$A=2\, \, \text{and}\, \, -A+B=1$$ which give B = 3.
Therefore $$\frac{2x+1}{(x-1)^2} = \frac{2}{x-1} + \frac{3}{(x-1)^2}.$$ Thus $$\int \frac{2x+1}{(x-1)^2}\,dx = \frac{2}{x-1}\,dx + \frac{3}{(x-1)^2}\,dx$$ $$= 2\, ln|x-1| - \frac{3}{x-1} + C.$$
$$\int \frac{2x+1}{(x-1)^2} \,dx.$$ We can rewrite as $$ \frac{2x+1}{(x-1)^2} = \frac{A}{x-1} + \frac{B}{(x-1)^2}$$ and so $$ \frac{2x+1}{(x-1)^2} = \frac{A(x-1)+B}{(x-1)^2}$$ Therefore, $$2x+1 = A(x-1) + B,$$ $$= Ax + (-A+B).$$ Equating the coefficients of corresponding powers of x, we get $$A=2\, \, \text{and}\, \, -A+B=1$$ which give B = 3.
Therefore $$\frac{2x+1}{(x-1)^2} = \frac{2}{x-1} + \frac{3}{(x-1)^2}.$$ Thus $$\int \frac{2x+1}{(x-1)^2}\,dx = \frac{2}{x-1}\,dx + \frac{3}{(x-1)^2}\,dx$$ $$= 2\, ln|x-1| - \frac{3}{x-1} + C.$$
Example 13. Find the function f(x) satisfying the equation
$$f'(x) = \frac{x-1}{\sqrt{x}} \, \, \text{with}\, \, f(1) = 0.$$
Solution
$$f'(x) = \frac{x-1}{\sqrt{x}}$$ By integrating we get $$f(x) = \int \frac{x-1}{\sqrt{x}}\,dx = \int (x^{\frac{1}{2}} -x^{-\frac{1}{2}})\,dx$$ $$= \frac{2}{3}x^{\frac{3}{2}} - 2x^{\frac{1}{2}} + C.$$ Since f(1) = 0, it follows that
$$f(1) = \frac{2}{3}(1)^{\frac{3}{2}} - 2(1)^{\frac{1}{2}} + C$$ $$0 = \frac{2}{3} - 2 + C $$ $$C = \frac{4}{3}.$$ Therefore $$f(x) = \frac{2}{3}x^{\frac{3}{2}} -2x^{\frac{1}{2}} + \frac{4}{3}.$$
$$f'(x) = \frac{x-1}{\sqrt{x}}$$ By integrating we get $$f(x) = \int \frac{x-1}{\sqrt{x}}\,dx = \int (x^{\frac{1}{2}} -x^{-\frac{1}{2}})\,dx$$ $$= \frac{2}{3}x^{\frac{3}{2}} - 2x^{\frac{1}{2}} + C.$$ Since f(1) = 0, it follows that
$$f(1) = \frac{2}{3}(1)^{\frac{3}{2}} - 2(1)^{\frac{1}{2}} + C$$ $$0 = \frac{2}{3} - 2 + C $$ $$C = \frac{4}{3}.$$ Therefore $$f(x) = \frac{2}{3}x^{\frac{3}{2}} -2x^{\frac{1}{2}} + \frac{4}{3}.$$
Example 14.
Find the function f(x) satisfying equation
$$f"(x) = \frac{e^x - e^{-x}}{2}, $$ with f'(0) = 1 and f(0) = 1.
Solution
$$f"(x) = \frac{e^x - e^{-x}}{2}$$ By integrating we get
$$f'(x) = \int \frac{e^x - e^{-x}}{2} \, dx = \frac{1}{2}(e^x+e^{-x}) + C.$$ Since f '(x) = 1, it follows that $$f'(0) = \frac{1}{2}(e^0+e^0)+C$$ $$1 = 1+ C$$ $$C=0.$$ Again, by integrating we get $$f(x) = \int\frac{e^x+e^{-x}}{2}\,dx=\frac{1}{2}(e^x-e^{-x})+C.$$ Since f(0) = 1, it follows that $$f(0) = \frac{1}{2}(e^0-e^0)+C$$ $$1=0+C$$ $$C=1.$$ Therefore, $$f(x) = \frac{1}{2}(e^x+e^{-x})+1.$$
$$f"(x) = \frac{e^x - e^{-x}}{2}, $$ with f'(0) = 1 and f(0) = 1.
Solution
$$f"(x) = \frac{e^x - e^{-x}}{2}$$ By integrating we get
$$f'(x) = \int \frac{e^x - e^{-x}}{2} \, dx = \frac{1}{2}(e^x+e^{-x}) + C.$$ Since f '(x) = 1, it follows that $$f'(0) = \frac{1}{2}(e^0+e^0)+C$$ $$1 = 1+ C$$ $$C=0.$$ Again, by integrating we get $$f(x) = \int\frac{e^x+e^{-x}}{2}\,dx=\frac{1}{2}(e^x-e^{-x})+C.$$ Since f(0) = 1, it follows that $$f(0) = \frac{1}{2}(e^0-e^0)+C$$ $$1=0+C$$ $$C=1.$$ Therefore, $$f(x) = \frac{1}{2}(e^x+e^{-x})+1.$$
- Use the partial fraction method to evaluate the following integrals. $$\text{(a)}\, \, \int \frac{1}{2x^2+5x+3}\,dx$$ $$\text{(b)\,}\, \int \frac{2x-1}{(x-3)^2} \, dx$$ $$\text{(c)\,}\, \int \frac{x+1}{(2x+5)(x+4)}\,dx$$ $$\text{(d)\,}\, \int \frac{2x^2-1}{x^2-1}\,dx$$
- Find the function f(x) that satisfying the equation $$f'(x) = sin\, 4x\, cos\,2x\, \, \text{with\,}\, f(\frac{\pi}{2})=0.$$
- Find the function g(x) that satisfying the equation $$g'(x) = x^2e^{x^3}\, \, \, \text{with\,}\, g(0) = -\frac{2}{3}.$$
- Find the function h(x) that satisfying the equation $$h'(x) = \frac{x}{x^2-1}\, \, \, \text{with\,}\, h(2) = \frac{1}{2}.$$
- Find the function f(x) that satisfying the equation $$f''(x)=2x-1\, \, \, \text{with\,}\, f(0)=-1\, \, \text{and\,}\, f'(1)=2.$$
- Find the function g(x) that satisfying the equation $$g''(x)=x\,sin\,x\, \, \, \text{with\,}\, g(\frac{\pi}{2})=0\, \, \text{and\,}\, g'(0)=0.$$
