Ask Questions, Get Answers

Want to ask us a question? Click here
Browse Questions
Home  >>  CBSE XII  >>  Math  >>  Integrals
0 votes

Evaluate the following: $\int e^{\tan^{-1}x}\bigg(\frac{1+x+x^2}{1+x^2}\bigg)dx$

Can you answer this question?

1 Answer

0 votes
  • Method of integration by parts
  • $\int udv=uv-\int vdu$
  • $1+\tan ^2 x=\sec^2 x$
Step 1:
Let $I=\large\int e^{\tan^{-1}x}\bigg(\large\frac{1+x+x^2}{1+x^2}\bigg)dx$
we can rewrite this
$I=\int\Large \frac{e^{\tan ^{-1}x}}{1+x^2}$$(1+x+x^2)dx$
Put $\tan ^{-1} x=t=>x=\tan t$
On differentiating with respect to t we get
Now substituting this we get,
$I=\int e^t(1+\tan t+\tan ^2 t)dt$
$=\int e^t(1+\tan ^2 t+\tan t)dt$
$1+\tan ^2 dt$
$=\int e^t(\sec ^2 t+\tan t)dt$
Now splitting the terms we get
$=\int e^t \sec ^2t dt+\int e^t. \tan t.dt$
Step 2:
Consider $\int e^t\; \tan t\; dt$
This is of the form $\int udv,$ hence this can be solved by the method of integration by parts.
$\int udv=uv-\int vdu$
Let $u=\tan t\qquad dv=e^t.dt$
$du=\sec^2 t\; dt \qquad v=e^t$
Hence $I=e^t.\tan t-\int e^t.\sec ^2 t.e^t.dt+\int e^t.\sec^2t.dt$
$=e^t.\tan t$
Substituting for t we get
$I=x.e^{\tan ^{-1}}x+c$
answered Apr 18, 2013 by meena.p
Ask Question
student study plans
JEE MAIN, CBSE, AIPMT Mobile and Tablet App
The ultimate mobile app to help you crack your examinations
Get the Android App