Question

$\int \frac{x^4+1}{x^6+1}dx={\tan }^{-1}{k}_1-\frac{2}{3}{\tan }^{-1}{k}_2+C,$ where

• A. $k_1=x+\frac{1}{x}$
• B. $k_2=x^3$
• C. $k_1=x-\frac{1}{x}$
• D. $k_2=x^4$

Answer 1

Answer: (B), (C)

The correct options are
B $k_2=x^3$
C $k_1=x-\frac{1}{x}$

$I=\int \frac{x^4+1}{x^6+1}dx=\int \frac{{(x^2+1)}^2-2x^2}{(x^2+1)(x^4-x^2+1)}dx$
$=\int \frac{\left(1+\frac{1}{x^2}\right)}{(x^2+\frac{1}{x^2}-1)}dx-2\int \frac{x^2}{{\left(x^3\right)}^2+1}dx$
$=\int \frac{\left(1+\frac{1}{x^2}\right)}{{(x-\frac{1}{x})}^2+1}dx-2\int \frac{x^2}{{\left(x^3\right)}^2+1}dx$
$[$In the first integral put $x-\frac{1}{x}=t$ and in the second put $x^3=u$$]$
$=\int \frac{dt}{t^2+1}-\frac{2}{3}\int \frac{du}{u^2+1}$
$={\tan }^{-1}t-\frac{2}{3}{\tan }^{-1}u+C$ $={\tan }^{-1}(x-\frac{1}{x})-\frac{2}{3}{\tan }^{-1}\left(x^3\right)+C$
$\therefore k_1=x-\frac{1}{x}$ and $k_2=x^3$