Question

Evaluate the following integral

$\int \frac{ {x}^{2} + 1 }{ {(x + 1)}^{2} } {e}^{x} dx$
​

Answer 1

$\large\underline{\sf{Solution-}}$

Given integral is

$\rm \: \displaystyle\int\tt \frac{ {x}^{2} + 1 }{ {(x + 1)}^{2} } {e}^{x} dx$

can be rewritten as

$\rm \: = \displaystyle\int\tt \frac{ {x}^{2} + 1 + 2x - 2x}{ {(x + 1)}^{2} } {e}^{x} dx$

can be re-arranged as

$\rm \: = \displaystyle\int\tt \frac{( {x}^{2} + 1 + 2x) - 2x}{ {(x + 1)}^{2} } {e}^{x} dx$

$\rm \: = \displaystyle\int\tt \frac{({x + 1)}^{2} - 2x}{ {(x + 1)}^{2} } {e}^{x} dx$

$\rm \: = \displaystyle\int\tt \bigg(1 - \frac{2x}{ {(x + 1)}^{2} }\bigg) {e}^{x} dx$

$\rm \: = \displaystyle\int\tt {e}^{x} \: dx \: - \: 2\displaystyle\int\tt \frac{x}{ {(x + 1)}^{2} } {e}^{x} \: dx$

$\rm \: = \tt {e}^{x} \: - \: 2\displaystyle\int\tt \frac{x + 1 - 1}{ {(x + 1)}^{2} } {e}^{x} \: dx$

$\rm \: = \tt {e}^{x} \: - \: 2\displaystyle\int\tt \bigg(\frac{x + 1}{ {(x + 1)}^{2} } - \frac{1}{ {(x + 1)}^{2} }\bigg) {e}^{x} \: dx$

$\rm \: = \tt {e}^{x} \: - \: 2\displaystyle\int\tt \bigg(\frac{1}{ {(x + 1)}} - \frac{1}{ {(x + 1)}^{2} }\bigg) {e}^{x} \: dx$

We know,

$\boxed{ \rm{ \:\displaystyle\int\tt {e}^{x}[ \: f(x) + f'(x) \: ] \: dx \: = \: {e}^{x} \: f(x) + c \: \: }}$

So, here

$\rm \: f(x) = \dfrac{1}{x + 1}$

and

$\rm \: f'(x) \: = \: - \: \dfrac{1}{(x + 1)^{2} }$

So, using above result, we get

$\tt \: = {e}^{x} - 2 \times {e}^{x} \times \dfrac{1}{x + 1} + c$

$\tt \: = {e}^{x} - \dfrac{2{e}^{x}}{x + 1} + c$

$\tt \: = {e}^{x}\bigg(1 - \dfrac{2}{x + 1}\bigg) + c$

$\tt \: = {e}^{x}\bigg(\dfrac{x + 1 - 2}{x + 1}\bigg) + c$

$\tt \: = {e}^{x}\bigg(\dfrac{x - 1}{x + 1}\bigg) + c$

Hence,

$\rm\implies \:\boxed{ \rm{ \:\tt \:\displaystyle\int\tt \frac{ {x}^{2} + 1}{ {(x + 1)}^{2} }{e}^{x}dx = {e}^{x}\bigg(\dfrac{x - 1}{x + 1}\bigg) + c \: }}$

$\rule{190pt}{2pt}$

Remark :-

Proof of the property :-

$\boxed{ \rm{ \:\displaystyle\int\tt {e}^{x}[ \: f(x) + f'(x) \: ] \: dx \: = \: {e}^{x} \: f(x) + c \: \: }}$

Consider

$\displaystyle\int\tt {e}^{x}[ \: f(x) + f'(x) \: ] \: dx$

$\rm \: = \displaystyle\int\tt {e}^{x}\: f(x) \: dx + \displaystyle\int\tt {e}^{x}f'(x) \: dx$

Now, using integration by parts in first integral, we get

$\rm \: = f(x)\displaystyle\int\tt {e}^{x} \: dx - \displaystyle\int\tt \bigg[\dfrac{d}{dx} f(x)\displaystyle\int\tt {e}^{x}dx\bigg] + \displaystyle\int\tt {e}^{x}f'(x) \: dx$

$\tt \: = f(x) {e}^{x} \: - \displaystyle\int\tt \: {e}^{x}f'(x) \: dx+ \displaystyle\int\tt {e}^{x}f'(x) \: dx + c$

$\tt \: ={e}^{x} f(x) \: + c$

Hence,

$\rm\implies \:\boxed{ \rm{ \:\displaystyle\int\tt {e}^{x}[ \: f(x) + f'(x) \: ] \: dx = {e}^{x}f(x) + c \: }}\:$