Question

If the primitive of $\frac{1}{x^6+x^4}$ is $f\left(x\right)+\frac{1}{x}+g\left(x\right)x^{-3}+C$, then

• A. $f\left(x\right)=ta{n}^{-1}x$
• B. $f\left(x\right)=2ta{n}^{-1}x$
• C. $g\left(x\right)$ is a constant function
• D. $g\left(x\right)=-1/3$

Answer 1

Answer: (A), (C), (D)

$g\left(x\right)$ is a constant function
$g\left(x\right)=-1/3$
$f\left(x\right)=ta{n}^{-1}x$

$\frac{1}{x^6+x^4}=\frac{1}{x^4(x^2+1)}=\frac{1}{x^2}(\frac{1}{x^2}-\frac{1}{x^2+1})=\frac{1}{x^4}-\frac{1}{x^2}+\frac{1}{x^2+1}$
Hence the primitive of $\frac{1}{x^6+x^4}$ is
$\frac{-1}{3}{x}^{-3}+\frac{1}{x}+{\tan }^{-1}x+c$