logo

Ask Questions, Get Answers

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

Integrate the function\[\tan^{-1}x\]

$\begin{array}{1 1} x\tan^{-1} x- \frac{1}{2} \log|1+x^2|+c \\ x\tan^{-1}x+ \frac{1}{2} \log|1+x^2|+c \\ x\tan^{-1}x- \frac{1}{2} \log|1-x^2|+c \\ x\tan^{-1}x+ \frac{1}{2} \log|1-x^2|+c \end{array} $

Can you answer this question?
 
 

1 Answer

0 votes
Toolbox:
  • (i)When there are two functions u and v and if they are of the form $\int u dv,$then we can solve it by the method of integration by parts\[\int udv=uv-\int vdu\]
  • (ii)$\;\frac{d}{dx}\tan^{-1}x=\frac{1}{1+x^2}.$
  • (iii)Method of substitution :If f(x)=t,then f'(x)dx=dt.
  • Hence $\int f(x)=\int t.dt+c.$
Given $ I=\int \tan^{-1}xdx.$
 
Clearly the given integral function is of the form $\int u dv$
 
Let $u=\tan^{-1}x.$Hence on differentiating we get,
 
du=$\frac{1}{1+x^2}dx.$
 
Let dv=dx,on integrating we get
 
v=x.
 
Now substituting for u,v,du and dv we get,
 
$\int \tan^{-1}xdx=x\tan^{-1}x-\int x.\frac{1}{1+x^2}dx.$
 
Let $I_1=\frac{x}{1+x^2}dx.$
 
Put $1+x^2=t.$
 
On differentiating with respect to x,
 
2xdx=dt.
 
xdx=$\frac{dt}{2}.$
 
Now substituting for t and dt we get,
 
$\frac{dt/2}{t}=\frac{1}{2}\int \frac{dt}{t}=\frac{1}{2}log|t|.$
 
Substituting for t we get
 
$\qquad=\frac{1}{2}log|1+x^2|.$
 
Now substituting for $I_1$ we get $I=x\tan^{-1}x-\frac{1}{2}log|1+x^2|+c.$

 

answered Feb 11, 2013 by sreemathi.v
 
Ask Question
student study plans
x
JEE MAIN, CBSE, NEET Mobile and Tablet App
The ultimate mobile app to help you crack your examinations
Get the Android App
...