Question

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

• A. both $f\left(x\right)$ and $g\left(x\right)$ are odd function
• B. $f\left(x\right)$ is monotonic function in $(0,\infty )$
• C. $f\left(x\right)=g\left(x\right)$ has no real roots
• D. $\int \frac{f\left(x\right)}{g\left(x\right)}dx=-\frac{1}{x}-\frac{1}{3x^3}+C$

Answer 1

Answer: (A), (B), (C)

$f\left(x\right)=g\left(x\right)$ has no real roots

Let $I=\int \frac{x^4+1}{x^6+1}dx$
$\Rightarrow I=\int \frac{(x^2+1{)}^2-2x^2}{(x^2+1)(x^4-x^2+1)}dx$
$\Rightarrow I=\int \frac{(1+\frac{1}{x^2})}{(x^2-1+\frac{1}{x^2})}dx-2\int \frac{x^2}{(x^3{)}^2+1}dx\Rightarrow I=\int \frac{(1+\frac{1}{x^2})}{{(x-\frac{1}{x})}^2+1}dx-2\int \frac{x^2}{(x^3{)}^2+1}dx$
Put $x-\frac{1}{x}=t\Rightarrow (1+\frac{1}{x^2})dx=dt$
And $x^3=u\Rightarrow 3x^{2}dx=du$
$\therefore I={\tan }^{-1}t-\frac{2}{3}{\tan }^{-1}u+C\Rightarrow I={\tan }^{-1}(x-\frac{1}{x})-\frac{2}{3}{\tan }^{-1}\left(x^3\right)+C$
$\therefore f\left(x\right)=x-\frac{1}{x},g\left(x\right)=x^3$
Clearly, both $f\left(x\right)$ and $g\left(x\right)$ are odd functions.

Also, $f^{\prime }\left(x\right)=1+\frac{1}{x^2}\ne 0$
and $f^{\prime }\left(x\right)>0\forall x\in (0,\infty )$
$g^{\prime }\left(x\right)=3x^2\ge 0$
Therefore, both $f\left(x\right)$ and $g\left(x\right)$ are monotonic.

$f\left(x\right)=g\left(x\right)\Rightarrow x-\frac{1}{x}=x^3\Rightarrow 1=\frac{1}{x^2}+x^2$
We know that,
$\frac{1}{x^2}+x^2\ge 2$
Hence no solution.

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