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

Evaluate the following:$\int\frac{x^2dx}{(x^2+a^2)(x^2+b^2)}$

$\begin{array}{1 1} (A)\;\large\frac{1}{a^2-b^2}\bigg[ a \;\tan ^{-1}(\large\frac{x}{a})+b\;\tan ^{-1} \large(\frac{x}{b})\bigg]+c \\(B)\;\large\frac{1}{a^2+b^2}\bigg[ a \;\tan ^{-1}(\large\frac{x}{a})- b\;\tan ^{-1} \large(\frac{x}{b})\bigg]+c \\ (C)\;\large\frac{1}{a^2-b^2}\bigg[ a \;\tan ^{-1}(\large\frac{x}{a})- b\;\tan ^{-1} \large(\frac{x}{b})\bigg]+c \\ (D)\;\large\frac{1}{a+b}\bigg[ a \;\tan ^{-1}(\large\frac{x}{a})- b\;\tan ^{-1} \large(\frac{x}{b})\bigg]+c \end{array} $

Can you answer this question?
 
 

1 Answer

0 votes
Toolbox:
  • A rational expression of the form $\large\frac{x}{(x+a)(x+b)}$ can be resolved in to partial fractions as $\large\frac{A}{x+a}+\frac{B}{x+b}$
  • $\int \large\frac{dx}{x^2+a^2}=\frac{1}{a} $$\tan ^{-1} \bigg(\large\frac{x}{a}\bigg)+c$
Let $I=\large\int\frac{x^2dx}{(x^2+a^2)(x^2+b^2)}$
$\large\frac{x^2}{(x^2+a^2)(x^2+b^2)}$ can be resolved into partial fractions as
$\large\frac{x^2}{(x^2+a^2)(x^2+b^2)}=\frac{Ax+B}{x^2+a^2}+\frac{Cx+D}{x^2+b^2}$
Now let us evaluate the values for A,B,C and D
$x^2=(Ax+B)(x^2+b^2)+(Cx+D)(x^2+a^2)$
Now equating the coefficient of like terms first equating for $x^3$
$0=A+C-------(1) =>A =-C$
Next equating for $x^2$
$1=B+D-----(2) =>B=1-D$
Next equating for $x$
$0=Ab^2+Ca^2-----(3)$
Next equating for constant term
$0=Bb^2+Da^2-----(4)$
Substituting for A in equ (3)
$-Cb^2+Ca^2=0$
$C(a^2-b^2)=0=>C=0$
Therefore $A=0$
Substituting for B in equ (4)
$(1-0)b^2+Da^2=0$
$b^2-Db^2+Da^2=0$
$D(a^2-b^2)=-b^2$
$=>\large\frac{-b^2}{a^2-b^2}$
Now $1-D=B$ substituting for D we get
$1+\large\frac{b^2}{a^2-b^2}=B$
Therefore $B=\large\frac{a^2-b^2+b^2}{a^2-b^2}=\frac{a^2}{a^2-b^2}$
Therefore $\large\frac{x^2}{(x^2+a^2)(x^2+b^2)}=\large\frac{a^2}{(a^2-b^2)(x^2+a^2)}-\large\frac{b^2}{(a^2-b^2)(x^2+b^2)}$
Hence $I=\large \frac{a^2}{a^2-b^2} \int \frac{dx}{x^2+a^2}-\frac{b^2}{a^2-b^2}\int \frac{dx}{x^2+b^2}$
This is of the form $\int \large\frac{dx}{x^2+a^2}=\frac{1}{a} $$\tan ^{-1}\large(\frac{x}{a})+c$
On integration we get
$I=\large\frac{a^2}{a^2-b^2}.\frac{1}{a} $$\tan ^{-1}(\large\frac{x}{a})-\frac{b^2}{a^2-b^2} \times \frac{1}{b} $$\tan ^{-1} \large(\frac{x}{b})+c$
$=\large\frac{a}{a^2-b^2}$$\tan ^{-1}(\large\frac{x}{a})-\frac{b}{a^2-b^2} $$\tan ^{-1} \large(\frac{x}{b})+c$
$=\large\frac{1}{a^2-b^2}\bigg[ $$a \;\tan ^{-1}(\large\frac{x}{a})- $$b\;\tan ^{-1} \large(\frac{x}{b})\bigg]+c$
answered Apr 22, 2013 by meena.p
 
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
...