Question

The value of $\int \frac{1+x^4}{1+x^6}dx$ is

• A. $ta{n}^{-1}x+ta{n}^{-1}{x}^3+C$
• B. $ta{n}^{-1}x+\frac{1}{3}ta{n}^{-1}{x}^3+C$
• C. $ta{n}^{-1}x-\frac{1}{3}ta{n}^{-1}{x}^3+C$
• D. $ta{n}^{-1}x+\frac{1}{3}ta{n}^{-1}{x}^2+C$

Answer 1

Answer: (B)

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