logo

Ask Questions, Get Answers

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

Evaluate : $ \int \large\frac{x^4dx}{(x-1)(x^2+1)} $

Can you answer this question?
 
 

1 Answer

0 votes
Toolbox:
  • $\int \large\frac{dx}{x^2+1}$$=\tan^{-1}x+c$
Step 1:
Let $I=\int \large\frac{x^4dx}{(x-1)(x^2+1)}$
$\large\frac{x^4}{(x-1)(x^2+1)}=\frac{x^4}{x^3+x-x^2-1}$
This is an improper rational function
$\Rightarrow (x+1)-\large\frac{1}{(x^3-x^2+x-1)}$
$\Rightarrow x+1-\large\frac{1}{(x-1)(x^2+1)}$
Step 2:
Consider $\large\frac{1}{(x-1)(x^2+1)}$
This can be resolved into partial fractions as
$\large\frac{A}{(x-1)}+\frac{Bx+C}{(x^2+1)}$
$1=A(x^2+1)+(x-1)(Bx+C)$
Put $x=1$
$1=2A\Rightarrow A=\large\frac{1}{2}$
Comparing the coefficient of $x^2$ term
$0=A+B\Rightarrow B=-\large\frac{1}{2}$
Comparing the coefficient of constant term
$1=A-C$
$C=-\large\frac{1}{2}$
Step 3:
$\therefore I=\int (x+1)dx-\int\large\frac{dx}{2(x-1)}+\int \frac{-1/2x-1/2}{x^2+1}$$dx-\large\frac{1}{2}\int \large\frac{dx}{x^2+1}$
On integrating we get,
$I=\large\frac{x^2}{2}$$+x-\large\frac{1}{2}$$\log(x-1)-\large\frac{1}{2}\times \frac{1}{2}$$\log (x^2+1)-\large\frac{1}{2}$$\tan^{-1}(x)+C$
$\;\;=\large\frac{x^2}{2}$$+x-\large\frac{1}{2}$$\log(x-1)-\large\frac{1}{4}$$\log(x^2+1)-\large\frac{1}{2}$$\tan^{-1}x+C$
$\;\;=\large\frac{x^2}{2}$$+x-\large\frac{1}{4}\frac{\log(x-1)^2}{x^2+1}-\large\frac{1}{2}$$\tan^{-1}x+C$
answered Nov 8, 2013 by sreemathi.v
 

Related questions

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
...