Question

If $\int \frac{x^4+1}{x^6+1}dx={\tan }^{-1}f\left(x\right)-\frac{2}{3}{\tan }^{-1}g\left(x\right)+C$ for constant of integration $C,$ then $f\left(x\right)-g\left(x\right)$ is

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

Answer 1

The correct option is A $x-\frac{1}{x}-x^3$

$I=\int \frac{x^4+1}{x^6+1}dx$
$=\int \frac{(x^2+1{)}^2-2x^2}{(x^2+1)(x^4-x^2+1)}dx$
$=\int \frac{(x^2+1)}{(x^4-x^2+1)}dx-2\int \frac{x^2}{x^6+1}dx$
$=\int \frac{(1+\frac{1}{x^2})}{x^2-1+\frac{1}{x^2}}dx-2\int \frac{x^2}{(x^3{)}^2+1}dx$

In the first integral, put $x-\frac{1}{x}=t$
$\Rightarrow (1+\frac{1}{x^2})dx=dt$
and in the second integral, put $x^3=u$
$\Rightarrow 3x^{2}dx=du$

$\therefore I=\int \frac{dt}{1+t^2}-\frac{2}{3}\int \frac{du}{1+u^2}$
$={\tan }^{-1}t-\frac{2}{3}{\tan }^{-1}u+C$
$={\tan }^{-1}(x-\frac{1}{x})-\frac{2}{3}{\tan }^{-1}\left(x^3\right)+C$

Here, $f\left(x\right)=x-\frac{1}{x}$ and $g\left(x\right)=x^3$