Question

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

• A. $0$
• B. $1$
• C. $3$
• D. $4$

Answer 1

The correct option is A $0$

$\int \frac{x^4+1}{x^6+1}dx={\tan }^{-1}\left(f\right(x\left)\right)-\frac{2}{3}{\tan }^{-1}\left(g\right(x\left)\right)+c$
Let
$I=\int \frac{(x^2+1{)}^2-2x^2}{(x^2+1)(x^4-x^2+1)}dx\Rightarrow I=\int \frac{(x^2+1)dx}{(x^4-x^2+1)}-2\int \frac{x^{2}dx}{(x^6+1)}\Rightarrow I=\int \frac{(1+\frac{1}{x^2})dx}{(x^2-1+\frac{1}{x^2})}-2\int \frac{x^{2}dx}{(x^3{)}^2+1}\Rightarrow I=\int \frac{(1+\frac{1}{x^2})dx}{1+{(x-\frac{1}{x})}^2}-2\int \frac{x^{2}dx}{(x^3{)}^2+1}$
Assuming $I_1=\int \frac{(1+\frac{1}{x^2})dx}{1+{(x-\frac{1}{x})}^2}$
Let $(x-\frac{1}{x})=t$
$\Rightarrow I_1=\int \frac{1}{1+t^2}dt\Rightarrow I_1={\tan }^{-1}t={\tan }^{-1}(x-\frac{1}{x})$
$I_2=\int \frac{x^{2}dx}{(x^3{)}^2+1}$
Let $x^3=t$
$\Rightarrow I_2=\frac{1}{3}{\tan }^{-1}\left(x^3\right)$
So,
$I={\tan }^{-1}(x-\frac{1}{x})-\frac{2}{3}{\tan }^{-1}\left(x^3\right)+cf\left(x\right)=x-\frac{1}{x},g\left(x\right)=x^3$
$f\left(x\right)$ and $g\left(x\right)$ are odd functions.

When $f\left(x\right)=g\left(x\right)$, then
$x-\frac{1}{x}=x^3\Rightarrow x^4-x^2+1=0\Rightarrow {(x^2-\frac{1}{2})}^2+\frac{3}{4}=0$
So, no real roots.