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Home  >>  CBSE XII  >>  Math  >>  Integrals
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Integrate the function\[\frac{5x}{(x+1)(x^2+9)}\]

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Toolbox:
  • If the function is of a rational form $\frac{1}{(x+a)(x^2+b)}$ then it can be resolved into its partial of the form $\frac{A}{x+a}+\frac{Bx+C}{x^2+b}$
  • (ii)$\int\frac{dx}{x+a}=log|x+a|+c$
Given:$I=\frac{5x}{(x+1)(x^2+9)}$
 
The given rational function is a proper rational function, and it can be resolved as
 
$\large\frac{5x}{(x+a)(x^2+9)}=\frac{A}{(x+1)}+\frac{Bx+c}{x^2+9}$
 
$=>5x=A((x^2+9)+(Bx+c)(x+1)$
 
Equating the coefficient of $x^2$.
 
$0=A+B$-----(1)$
 
Equating the coefficient of x.
 
$5=B+C$-----(2)
 
Equating the constant terms,
 
$0=9A+C$-----(3)
 
Subtract equ(1) from equ (2)
 
$A+B=0$
$B+C=5$
___________
$A-C=-5$-----(4)
 
Add equ (4) and equ(3)
 
$9A+C=0$
$A-C=-5$
_______________
$10A=-5$
 
$=>A=-\frac{1}{2}$
 
Substituting for A in equ(1) we get
 
$-\frac{1}{2}+B=0\qquad =>B=\frac{1}{2}$
 
Substituting for B in equ(2) we get
 
$B+C=5$
 
(ie) $\frac{1}{2}+c=5 \qquad=>c=5-\frac{1}{2}$
 
$=\frac{9}{2}$
 
Hence $A=-\frac{1}{2},B=\frac{1}{2}\;and\; C=\frac{9}{2}$
 
Now substituting for A,B,C
 
$\large\frac{5x}{(x+1)(x^2+9)}=-\frac{1}{2(x+1)}+\frac{1/2x+9/2}{(x^2+9)}$
 
$=-\large\frac{1}{2(x+1)}+\frac{x}{2(x^2+9)}+\frac{9}{2(x^2+9)}$
 
Therefore $I=-\large\frac{1}{2}\int \frac{dx}{x+1}+\frac{1}{2}\int\frac {xdx}{x^2+9}+\frac{9}{2}\int \frac{dx}{x^2+9}$
 
consider $\int \frac{x}{x^2+9}dx$
 
Let $x^2+9=t$
 
Consider $\int \frac{dx}{x^2+9}$ which is of the form $\int \frac{dx}{x^2+a^2}=\frac{1}{a}\tan ^{-1}(x/a)$
 
$\frac{1}{3} \tan^{-1}(x/3)$
 
On differentiating we get,
 
$2xdx=dt$
 
$=>xdx=dt/2$
 
Therefore $\int \frac{dt/2}{t}=\frac{1}{2} log |x^2+9|$
 
On integrating I we get
 
$-\frac{1}{2}log(x+1)+\frac{1}{2}(\frac{1}{2} log(x^2+9)|+\frac{9}{2} \times \frac{1}{3} \tan ^{-1}(x/3)$
 
$-\large\frac{1}{2}log|x+1|+\frac{1}{4}log|x^2+9|+\frac{3}{2} \tan ^{-1}(x/3)$

 

 

answered Feb 15, 2013 by meena.p
 
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