Question

The value of the integral $\int \frac{(x^2+1)(x^2+2)}{(x^2+3)(x^2+4)}dx$ is
(where $C$ is integration constant)

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

Answer 1

The correct option is C $x+\frac{2}{\sqrt{3}}{\tan }^{-1}\frac{x}{\sqrt{3}}-3{\tan }^{-1}\frac{x}{2}+C$

$\frac{(x^2+1)(x^2+2)}{(x^2+3)(x^2+4)}=1-\frac{4x^2+10}{(x^2+3)(x^2+4)}$
Let $x^2=t$
$\frac{(x^2+1)(x^2+2)}{(x^2+3)(x^2+4)}=1-\frac{4t+10}{(t+3)(t+4)}$
Now, assuming
$\frac{4t+10}{(t+3)(t+4)}=\frac{A}{t+3}+\frac{B}{t+4}\Rightarrow 4t+10=A(t+4)+B(t+3)$
Equating the coefficients of $t$ and constant, we obtain
$A+B=44A+3B=10\Rightarrow A=-2;B=6$
Now,
$\Rightarrow 1-\frac{4t+10}{(t+3)(t+4)}=1+\frac{2}{t+3}-\frac{6}{t+4}\Rightarrow \int \frac{(x^2+1)(x^2+2)}{(x^2+3)(x^2+4)}dx=\int [1+\frac{2}{(x^2+3)}-\frac{6}{(x^2+4)}]dx=\int [1+\frac{2}{x^2+{\left(\sqrt{3}\right)}^2}-\frac{6}{x^2+2^2}]dx=x+2\left(\frac{1}{\sqrt{3}}{\tan }^{-1}\frac{x}{\sqrt{3}}\right)-6\left(\frac{1}{2}{\tan }^{-1}\frac{x}{2}\right)+C=x+\frac{2}{\sqrt{3}}{\tan }^{-1}\frac{x}{\sqrt{3}}-3{\tan }^{-1}\frac{x}{2}+C$