# Methods of Integration: Integration by Parts

#### notes

If u and v are any two differentiable functions of a single variable x . Then, by the product rule of differentiation, we have
d/(dx)(uv) = u(dv)/(dx) + v(du)/(dx)
Integrating both sides, we get

uv = int u (dv)/(dx) dx + int v (du)/(dx) dx

or int u (dv)/(dx)dx = uv - int v (du)/(dx) dx        ...(1)

Let u = f(x) and (dv)/(dx)=g(x) . Then

(du)/(dx) = f'(x) and v = int g(x) dx

Therefore, expression (1) can be rewritten as
int f(x) g(x) dx = f(x)int g(x) dx - int [ int g(x) dx] f'(x) dx

i.e. int f(x) g(x) dx = f(x) int g(x) dx - int [f'(x) int g(x) dx] dx

If we take f as the first function and g as the second function, then this formula may be stated as follows:

“The integral of the product of two functions
= (first function) × (integral of the second function) – Integral of [(differential coefficient of the first function) × (integral of the second function)]”

#### description

• int(u.v) dx = u intv dx - int((du)/(dx)).(intvdx) dx
• Integral of the type ∫ ex [ f(x) + f'(x)] dx = exf(x) + C
• Integrals of some more types
1. I = int sqrt(x^2 - a^2) dx = x/2 sqrt(x^2 - a^2) - a^2/2 log | x + sqrt(x^2 - a^2 )| + C
2. I = int sqrt(x^2 - a^2) dx = x/2 sqrt(x^2 - a^2) - a^2/2 log | x + sqrt(x^2 - a^2 )| + C
3. I = int sqrt(x^2 - a^2) dx = x/2 sqrt(x^2 - a^2) - a^2/2 log | x + sqrt(x^2 - a^2 )| + C
If you would like to contribute notes or other learning material, please submit them using the button below.

#### Video Tutorials

We have provided more than 1 series of video tutorials for some topics to help you get a better understanding of the topic.

Series 1

Series 2

### Shaalaa.com

Integrals part 30 (Integration by parts) [00:13:59]
S
Share