Question

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

• A. $-\ln |\frac{2x^6+x^4+1}{x^4}|+c$
• B. $\frac{-x^4}{2x^6+x^4+1}+c$
• C. $\frac{-x^2}{2x^6+x^4+1}+c$
• D. $\frac{-1}{2x^6+x^4+1}+c$

Answer 1

The correct option is B $\frac{-x^4}{2x^6+x^4+1}+c$

$\int \frac{4x^3(x^6-1)}{(2x^6+x^4+1{)}^2}dx$
$=\int \frac{4x^9-4x^3}{x^8(2x^2+1+x^{-4}{)}^2}dx=\int \frac{(4x-4x^{-5})}{(2x^2+1+x^{-4}{)}^2}dx$
$2x^2+1+x^{-4}=t$
$\Rightarrow (4x-4x^{-5})dx=dt$
$\int \frac{dt}{t^2}=-\frac{1}{t}+c$
$=-\frac{1}{(2x^2+1+x^{-4})}+c=\frac{-x^4}{2x^6+x^4+1}+c$