Question

$\int \frac{dx}{x\sqrt{x^6-16}}=$

• A. $\sec \frac{x^3}{4}+c$
• B. $\frac{1}{12}\sec \frac{x^3}{4}+c$
• C. $\frac{1}{3}\sec \frac{x^3}{4}+c$
• D. $Noneofthese$

Answer 1

The correct option is B $\frac{1}{12}\sec \frac{x^3}{4}+c$

let,

$\begin{array}{c}{x}^3=t\\ 3x^{2}dx=dt\\ dx=\frac{dt}{3x^2}\\ So,\\ I=\int \frac{dt}{3x^2}\times \frac{1}{x\sqrt{{\left(x^3\right)}^2-16}}\\ =\int \frac{dt}{3x^3\sqrt{{\left(t\right)}^2-16}}\\ =\frac{1}{3}\int \frac{dt}{t\sqrt{{\left(t\right)}^2-4^2}}\\ As,\\ \int \frac{dx}{x\sqrt{{\left(x\right)}^2-a^2}}=\frac{1}{a}{\sec }^{-1}\left(\frac{x}{a}\right)\\ So,\\ I=\frac{1}{3}\times \frac{1}{4}{\sec }^{-1}\left(\frac{t}{4}\right)+C\\ =\frac{1}{12}{\sec }^{-1}\left(\frac{x^3}{4}\right)+C\end{array}$

Thus Option B