Question

$\int \frac{1}{x^6+x^4}dx$ is equal to

• A. $-\frac{1}{3x^3}+\frac{1}{x}+\cos \text{e}{\text{c}}^{-1}x+C$
• B. $-\frac{1}{3x^3}+\frac{1}{x}+{\cot }^{-1}x+C$
• C. $-\frac{1}{3x^3}+\frac{1}{x}+ta{n}^{-1}x+C$
• D. None of the above

Answer 1

Answer: (C)

Let $I=\int \frac{1}{x^6+x^4}dx=\int \frac{1}{x^4(x^2+1)}dx$
On putting $x=\tan \theta$ and $dx={\sec }^2\theta d\theta$ ,
we get $I=\int \frac{1}{{\tan }^4\theta (1+{\tan }^2\theta )}{\sec }^2\theta d\theta$
$=\int {\cot }^4\theta d\theta$
$=\int {\cot }^2\theta (\cos \text{e}{\text{c}}^2\theta -1)d\theta$
$=\int {\cot }^2\theta \cos \text{e}{\text{c}}^2\theta d\theta -\int {\cot }^2\theta d\theta$
$=-\int {\cot }^2\theta d(\cot \theta )-\int (\cos \text{e}{\text{c}}^2\theta -1)d\theta$
$=-\frac{{\cot }^3\theta }{3}+\cot \theta +\theta +C$
$=-\frac{1}{3x^3}+\frac{1}{x}+{\tan }^{-1}+C$