Mathematics

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

##### SOLUTION
$\int e^{tan^{-1}x}(\frac{1+x+x^{2}}{1+x^{2}})dx$
put $tan^{-1}x=t$
x= tan t
$dx = sec^{2}t$ $dt$
$\int e^{t}(\frac{1+ tan t+ tan^{2}t}{1+tan^{2}t})sec^{2}t dt$
$(sec^{2}x- tan^{2}x=1 \Rightarrow 1+ tan^{2}x = sec^{2}x)$
$\int e^{t}(\frac{tan t + sec^{2}t}{sec^{2}t})sec^{2}t dt$
$\int e^{t}(\frac{tan t sec^{2}t}{sec^{2}t}+\frac{sec^{2}t sec^{2}t}{sec^{2}t})dt$
$\int e^{t}(tan t + sec^{2}t)dt$
$\int e^{t} tan t dt + \int e^{t} sec^{2}t dt$
Integrate by parts $I_{1}$
$tan t e^{t}- \int sec^{2}t \int e^{t}dt + \int e^{t} sec^{2} t dt$
$tan t e^{t} - \int sec^{2}t \int e^{t} dt + \int e^{t} sec^{2} t dt$
$= e^{t} tan^{t}+C$
Now put $t= tan^{-1}x$
$= e^{tan^{-1}x} tan x (tan^{-1}x)+C$
$= e^{tan^{-1}x}+C$
$= xe^{tan^{-1}x}+C$

Its FREE, you're just one step away

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

#### Realted Questions

Q1 Subjective Medium
$\displaystyle \int \sin \theta \cos \theta d\theta$

1 Verified Answer | Published on 17th 09, 2020

Q2 Subjective Medium
Evaluate the following integral:
$\int { \cfrac { \sin { \left( \log { x } \right) } }{ x } } dx\quad$

1 Verified Answer | Published on 17th 09, 2020

Q3 Single Correct Medium
If $f(x)=\displaystyle \int_0^{x}t \sin t dt$, then $f'(x)$ is
• A. $\cos x + x \sin x$
• B. $x \cos x$
• C. $\sin x + x \cos x$
• D. $x \sin x$

1 Verified Answer | Published on 17th 09, 2020

Q4 Single Correct Medium
$\displaystyle \int_{\sin \theta}^{\cos \theta} f(x \tan \theta) d x\left(\text { where } \theta \neq \dfrac{n \pi}{2}, n \in I\right)$ is equal to
• A. $-\tan \theta \int_{\cos \theta}^{\sin \theta} f(x) d x$
• B. $\sin \theta \int_{1}^{\tan \theta} f(x \cos \theta) d x$
• C. $\frac{1}{\tan \theta} \int_{\sin \theta}^{\sin \theta \tan \theta} f(x) d x$
• D. $-\cos \theta \int_{1}^{\tan \theta} f(x \sin \theta) d x$

$\displaystyle\int \dfrac{\log x}{x}dx$