Question

$\int \frac{x^2}{(x^2+1)(x^2+4)}dx$ is equal to

• A.
  $\frac{-1}{3}{\tan }^{-1}x+\frac{2}{3}{\tan }^{-1}\frac{x}{2}+C$
• B.
  $\frac{-1}{3}{\tan }^{-1}x-\frac{2}{3}{\tan }^{-1}\frac{x}{2}+C$
• C.
  $\frac{1}{3}{\tan }^{-1}x+\frac{2}{3}{\tan }^{-1}\frac{x}{2}+C$
• D. $\frac{1}{3}{\tan }^{-1}x-\frac{2}{3}{\tan }^{-1}\frac{x}{2}+C$

Answer 1

The correct option is A

$\frac{-1}{3}{\tan }^{-1}x+\frac{2}{3}{\tan }^{-1}\frac{x}{2}+C$
These kind of questions can be solved easily using integration by partial fraction.
Here, we can write our integrand as
$\frac{x^2}{(x^2+1)(x^2+4)}=\frac{ax+b}{x^2+1}+\frac{cx+d}{x^2+4}$
$\Rightarrow \frac{x^2}{(x^2+1)(x^2+4)}=\frac{(ax+b)(x^2+4)+(cx+d)(x^2+1)}{(x^2+1)(x^2+4)}$
$\Rightarrow x^2=(ax+b)(x^2+4)+(cx+d)(x^2+1)\Rightarrow x^2=(a+c)x^3+(d+b)x^2+(4a+c)x+(4b+d)$
Now, equating the co-efficients of different powers of $x$ we get the equations:
$a+c=0;4a+c=0$
and
$b+d=1;4b+d=0$
Now solving these two sets of equations we get:
$a=0,c=0andb=\frac{-1}{3},d=\frac{4}{3}$
Now, we can write our integral as
$\int \frac{x^2}{(x^2+1)(x^2+4)}dx=\frac{1}{3}\int \left(\frac{4}{x^2+4}-\frac{1}{x^2+1}\right]dx$
we know that $\int \frac{1}{x^2+a^2}dx=\frac{1}{a}{\tan }^{-1}\left(\frac{x}{a}\right)+C$
So, we get integral as:
$I=\frac{-1}{3}{\tan }^{-1}\left(x\right)+\frac{4}{3}\times \frac{1}{2}{\tan }^{-1}\left(\frac{x}{2}\right)+C\Rightarrow I=-\frac{1}{3}{\tan }^{-1}\left(x\right)+\frac{2}{3}{\tan }^{-1}\left(\frac{x}{2}\right)+C$