Mathematics

# $\int { { e }^{ x } } { \left( \dfrac { x-1 }{ { x }^{ 2 }+1 } \right) }^{ 2 }dx$

$\dfrac { { e }^{ x } }{ { x }^{ 2 }+1 } +c$

##### SOLUTION
${\int {{e^x}\left( {\dfrac{{x - 1}}{{{x^2} + 1}}} \right)} ^2}dx$
$= \int {{e^x}\dfrac{{{{\left( {x - 1} \right)}^2}}}{{{{\left( {{x^2} + 1} \right)}^2}}}} dx$
$= \int {{e^2}\left[ {\dfrac{{\left( {{x^2} + 1} \right) - 2x}}{{{{\left( {{x^2} + 1} \right)}^2}}}} \right]}$
$= \int {{e^x}} \left[ {\dfrac{{{x^2} + 1}}{{{{\left( {{x^2} + 1} \right)}^2}}} - \dfrac{{2x}}{{{{\left( {{x^2} + 1} \right)}^2}}}} \right]$
$= \int {{e^x}} \left[ {\dfrac{1}{{{{\left( {{x^2} + 1} \right)}}}} - \dfrac{{2x}}{{{{\left( {{x^2} + 1} \right)}^2}}}} \right]$
$= \dfrac{{{e^x}}}{{{x^2} + 1}} + c$

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Single Correct Medium Published on 17th 09, 2020
Questions 203525
Subjects 9
Chapters 126
Enrolled Students 84

#### Realted Questions

Q1 Multiple Correct Hard
If $I_n = \displaystyle \int_{-\pi}^{\pi} \dfrac{sinnx}{(1+\pi^x)sinx}dx,$ n=0,1,2,..., then
• A. $I_n=I_{n+1}$
• B. $I_n=I_{n+2}$
• C. $\displaystyle \sum_{m=1}^{10} I_{2m+1}=10\pi$
• D. $\displaystyle \sum_{m=1}^{10} I_{2m}=0$

1 Verified Answer | Published on 17th 09, 2020

Q2 Single Correct Medium
The value of $\displaystyle\int\limits_{0}^{\frac{\pi}{4}} \tan^2 \theta\ d\theta=$
• A. $\dfrac{\pi}{4}-1$
• B. $\dfrac{\pi}{4}$
• C. none of these
• D. $1-\dfrac{\pi}{4}$

1 Verified Answer | Published on 17th 09, 2020

Q3 Single Correct Medium
$\int _{ 0 }^{ 2\pi }{ \sqrt { 1+\sin { \dfrac { x }{ 2 } } } dx } =$
• A. $2$
• B. $8$
• C. $4$
• D. $0$

1 Verified Answer | Published on 17th 09, 2020

Q4 Single Correct Hard
If $\displaystyle\int{\frac{e^x(2-x^2)}{(1-x)\sqrt{1-x^2}}dx}=\mu e^x{\left(\frac{1+x}{1-x}\right)}^{\lambda}+C$, then $2(\lambda+\mu)$ is equal to
• A. $-1$
• B. $0$
• C. $2$
• D. $3$

$\displaystyle\int \dfrac{1}{e^x+e^{-x}}dx$.