Question

If $\int \frac{x^4+1}{x^6+1}dx={\tan }^{-1}f\left(x\right)-\frac{2}{3}{\tan }^{-1}g\left(x\right)+c,$ where $c$ is an arbitrary constant, then

• A. both $f\left(x\right)$ and $g\left(x\right)$ are odd functions
• B. both $f\left(x\right)$ and $g\left(x\right)$ are even functions
• C. $f\left(x\right)=g\left(x\right)$ has no real roots
• D. $\int \frac{f\left(x\right)}{g\left(x\right)}dx=\frac{-1}{x}+\frac{1}{3x^3}+d,$ where $d$ is an arbitrary constant.

Answer 1

Answer: (A), (C), (D)

The correct options are
A both $f\left(x\right)$ and $g\left(x\right)$ are odd functions
C $f\left(x\right)=g\left(x\right)$ has no real roots
D $\int \frac{f\left(x\right)}{g\left(x\right)}dx=\frac{-1}{x}+\frac{1}{3x^3}+d,$ where $d$ is an arbitrary constant.

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

$I={\tan }^{-1}(x-\frac{1}{x})-\frac{2}{3}{\tan }^{-1}\left(x^3\right)+c$
So, $f\left(x\right)=x-\frac{1}{x}$ and $g\left(x\right)=x^3$

$\int \frac{f\left(x\right)}{g\left(x\right)}dx=\int \frac{(x-\frac{1}{x})}{x^3}dx$
$=\int (\frac{1}{x^2}-\frac{1}{x^4})dx$
$=-\frac{1}{x}+\frac{1}{3x^3}+d$