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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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  • 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)$\int\frac{dx}{x+a}=log|x+a|+c$
The given rational function is a proper rational function, and it can be resolved as
Equating the coefficient of $x^2$.
Equating the coefficient of x.
Equating the constant terms,
Subtract equ(1) from equ (2)
Add equ (4) and equ(3)
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
(ie) $\frac{1}{2}+c=5 \qquad=>c=5-\frac{1}{2}$
Hence $A=-\frac{1}{2},B=\frac{1}{2}\;and\; C=\frac{9}{2}$
Now substituting for A,B,C
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,
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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