Question

$\int \frac{dx}{\sqrt{x^{10}-x^2}},x>1=$ ____ $+C$

• A. $\frac{1}{4}\log |\sqrt{x^{10}-x^2}+x^2|$
• B. $\frac{1}{2}\log |x^{10}-x^2|$
• C. $-\frac{1}{4}{\sec }^{-1}\left(x^4\right)$
• D. $\frac{1}{4}{\sec }^{-1}\left(x^4\right)$

Answer 1

The correct option is D $\frac{1}{4}{\sec }^{-1}\left(x^4\right)$

$I=\int \frac{dx}{\sqrt{x^{10}-x^2}}=\int \frac{dx}{x\sqrt{x^8-1}}$

$=\int \frac{4x^{3}dx}{4x^4\sqrt{x^8-1}}$

Let $x^4=t\Rightarrow 4x^{3}dx=dt$
$I=\int \frac{dt}{4t\sqrt{t^2-1}}$
$=\frac{1}{4}{\sec }^{-1}\left(t\right)+C$
$=\frac{1}{4}{\sec }^{-1}\left(x^4\right)+C$