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

Answer: (D)

$\frac{1}{4}{\sec }^{-1}\left(x^4\right)$

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

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

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

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