Question

The value of the integral $\int \frac{x^4+1}{x^6+1}dx$ is
$(C$ is a constant of integration$)$

• A. ${\tan }^{-1}x+3{\tan }^{-1}{x}^3+C$
• B. $\frac{1}{3}{\tan }^{-1}x+{\tan }^{-1}{x}^3+C$
• C. $\frac{1}{3}{\tan }^{-1}x+3{\tan }^{-1}{x}^3+C$
• D. ${\tan }^{-1}x+\frac{1}{3}{\tan }^{-1}{x}^3+C$

Answer 1

The correct option is D ${\tan }^{-1}x+\frac{1}{3}{\tan }^{-1}{x}^3+C$

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