# Integration By Parts

Contents

### Summary

Remember the formula for ‘Integration By Parts’: $\int { \quad u\quad \frac { dv }{ dx } } \quad =\quad (u\quad v)\quad -\quad \int { \quad v\quad \frac { du }{ dx } }$

We now know how to evaluate many basic integrals. However there are many integrals which are in the form of two functions and cannot be simplified by any substitution. In such cases we use the ‘Product Rule’ of differentiation.

We know that: $\frac { d }{ dx } \quad (u\quad v)\quad =\quad u\quad \frac { dv }{ dx } \quad +\quad v\quad \frac { du }{ dx }$

Rewrite it as: $u\quad \frac { dv }{ dx } \quad =\quad \frac { d }{ dx } \quad (u\quad v)\quad -\quad v\quad \frac { du }{ dx }$

Integrating both sides: $\int { \quad u\quad \frac { dv }{ dx } } \quad =\quad (u\quad v)\quad -\quad \int { \quad v\quad \frac { du }{ dx } }$

We call the above equation the method of integration by parts.

The advantage of using the integration-by-parts formula is that we can use it to exchange the integrals, to be able to make it easier to solve the integral. The following example illustrates how its applied.

#### Example #1

Q.   Evaluate $\int { \quad xsin(x) } dx$

Solution:

Using the integration by parts formula: $\int { \quad u\quad \frac { dv }{ dx } } \quad =\quad (u\quad v)\quad -\quad \int { \quad v\quad \frac { du }{ dx } }$

Let: $u\quad =\quad x$   & $\frac { dv }{ dx } \quad =\quad sin(x)$

Then: $\frac { du }{ dx } \quad =\quad 1$   & $v\quad =\quad -cos(x)$

Put values in the above formula: $\int { xsin(x) } dx\quad =\quad x(-cos(x))\quad -\quad \int { (-cos(x)) } (1)dx$ $=\quad -xcos(x)\quad +\quad \int { cos(x) } dx$

Ans: $=\quad -xcos(x)\quad +\quad sin(x)\quad +\quad C$

#### Example #2

Q.   Integrate ${ x }^{ 2 }{ e }^{ x }$  with respect to x

Solution:

Let: $u\quad =\quad { x }^{ 2 }$     & $\frac { dv }{ dx } \quad =\quad { e }^{ x }$

Then: $\frac { du }{ dx } \quad =\quad { 2x }$     & $v\quad =\quad { e }^{ x }$

Hence, substituting values in the formula of integration by parts: $\int { { x }^{ 2 } } { e }^{ x }dx\quad =\quad ({ x }^{ 2 })({ e }^{ x })\quad -\quad \int { ({ e }^{ x }) } (2x)dx$ $\quad =\quad ({ x }^{ 2 })({ e }^{ x })\quad -\quad 2\int { ({ xe }^{ x }) }$

In this case we need to integrate the term on the right again: $\int { { x }^{ 2 } } { e }^{ x }dx\quad =\quad ({ x }^{ 2 })({ e }^{ x })\quad -\quad 2\left[ x{ e }^{ x }\quad -\quad { e }^{ x } \right] \quad +\quad C$

Ans: $=\quad ({ x }^{ 2 })({ e }^{ x })\quad -\quad 2x{ e }^{ x }\quad -\quad 2{ e }^{ x }\quad +\quad C$

#### Example #3

Q.   Evaluate $\int { \ln { (x) } dx }$

Solution:

We rewrite the above integral in the form: $\int { (1)\ln { (x) } dx }$

Now: $u\quad =\quad \ln { (x) }$     & $\frac { dv }{ dx } \quad =\quad 1$

Then: $\frac { du }{ dx } \quad =\quad \frac { 1 }{ x }$     & $v\quad =\quad x$

Hence, substituting values in the formula of integration by parts: $\int { \ln { (x) } dx } \quad =\quad (x)\ln { (x) } \quad -\quad \int { x\quad \frac { 1 }{ x } } \quad dx$ $=\quad (x)\ln { (x) } \quad -\quad \int { 1 } dx$

Ans: $=\quad x\ln { (x) } \quad -\quad x\quad +\quad C$