Question

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

• 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)

$\int \frac{f\left(x\right)}{g\left(x\right)}dx=-\frac{1}{x}+\frac{1}{3x^3}+d,$ where $d$ is an arbitrary constant

Let $I=\int \frac{(x^4+1)}{(x^6+1)}dx$
$\Rightarrow I=\int \frac{(x^2+1{)}^2-2x^2}{(x^2+1)(x^4-x^2+1)}dx$
$\Rightarrow I=\int \frac{(x^2+1)dx}{(x^4-x^2+1)}-2\int \frac{x^{2}dx}{(x^6+1)}$
$\Rightarrow I=\int \frac{(1+\frac{1}{x^2})dx}{(x^2-1+\frac{1}{x^2})}-2\int \frac{x^{2}dx}{(x^3{)}^2+1}$
$\Rightarrow I=I_1-I_2$ (say)

$I_1=\int \frac{(1+\frac{1}{x^2})dx}{(x^2-1+\frac{1}{x^2})}$
$\Rightarrow I_1=\int \frac{1+\frac{1}{x^2}}{{(x-\frac{1}{x})}^2+1}dx$
Put $x-\frac{1}{x}=t$
$\Rightarrow (1+\frac{1}{x^2})dx=dt$
$I_1=\int \frac{1}{t^2+1}dt$
$\Rightarrow I_1={\tan }^{-1}t+C_1$
$\therefore I_1={\tan }^{-1}(x-\frac{1}{x})+C_1$

and $I_2=2\int \frac{x^{2}dx}{(x^3{)}^2+1}$
Put $x^3=p\Rightarrow 3x^{2}dx=dp$
$I_2=\frac{2}{3}\int \frac{dp}{p^2+1}$
$\Rightarrow I_2=\frac{2}{3}{\tan }^{-1}p+C_2$
$\therefore I_2=\frac{2}{3}{\tan }^{-1}\left(x^3\right)+C_2$

$I={\tan }^{-1}(x-\frac{1}{x})-\frac{2}{3}{\tan }^{-1}\left(x^3\right)+c(c=C_1-C_2)$
$\therefore f\left(x\right)=x-\frac{1}{x}$ and $g\left(x\right)=x^3$


$f(-x)=-f\left(x\right)$ and $g(-x)=-g\left(x\right)$
$\therefore$ Both $f\left(x\right)$ and $g\left(x\right)$ are odd functions.


If $f\left(x\right)=g\left(x\right),$
$x^3=x-\frac{1}{x}(x\ne 0)$
$\Rightarrow x^4-x^2+1=0$
Let $x^2=t.$
Then, $t^2-t+1=0$
$D=1-4=-3<0$
Hence, $f\left(x\right)=g\left(x\right)$ has no real roots.


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