Mathematics

# The value of $\int { { e }^{ { \tan ^{ -1 } { x } } }} \dfrac { \left( 1+x{ +x }^{ 2 } \right) }{ 1{ +x }^{ 2 } } dx$ is ?

${ xe }^{ \tan ^{ -1 }{ x+C } }$

##### SOLUTION
$\displaystyle \tan^{-1} x = t \quad \Rightarrow \quad x = \tan t$

$\displaystyle dx. \frac{1}{1+x^2} = dt.$

$\displaystyle I = \int e^t (1+\tan t + \tan^2 t) dt$

$\displaystyle = \int e^t (\tan t+ \sec^2 t ) dt \quad \left[\because \int e^x (f(x) + f'(x). dx = e^x f(x) +c\right]$

$\displaystyle = e^t \tan t +c$

$\displaystyle = e^{\tan^{-1} x} \tan (\tan^{-1}x) + c$

$\displaystyle I = x \, e^{\tan^{-1}x} + c$

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Single Correct Medium Published on 17th 09, 2020
Questions 203525
Subjects 9
Chapters 126
Enrolled Students 86

#### Realted Questions

Q1 Single Correct Medium
If $\displaystyle I = \int \frac {dx}{\tan x \log \: cosec \: x}$, then I equals
• A. $\displaystyle \log |\log \: cosec \: x| + C$
• B. $\displaystyle \log |\log \cos x| + C$
• C. $\displaystyle - \log |\log (cosec \: x)| + C$
• D. none of these

1 Verified Answer | Published on 17th 09, 2020

Q2 Subjective Hard
Evaluate $\displaystyle\int\displaystyle\dfrac{1}{e^x-1}dx$.

1 Verified Answer | Published on 17th 09, 2020

Q3 Single Correct Medium

$\displaystyle \frac{1}{a^{2}-x^{2}}=$
• A. $\displaystyle \frac{1}{a(a-x)}+\frac{1}{2a(a+x)}$
• B. $\displaystyle \frac{1}{3a(a-x)}+\frac{1}{2a(a+x)}$
• C. $\displaystyle \frac{1}{2a(a-x)}+\frac{1}{a(a+x)}$
• D. $\displaystyle \frac{1}{2a(a-x)}+\frac{1}{2a(a+x)}$

1 Verified Answer | Published on 17th 09, 2020

Q4 Subjective Medium
Solve $\displaystyle\int \dfrac { \left( x ^ { 3 } + 8 \right) ( x - 1 ) } { x ^ { 2 } - 2 x + 4 }dx$

Solve $\displaystyle\int \dfrac{x}{{{{\left( {x + 1} \right)}^2}}}dx$