Question

$\int \sqrt{\frac{x^3}{x^5-1}}dx=$

• A. $\frac{2}{3}{\cosh }^{-1}\left(x^{5/2}\right)+c$
• B. $\frac{2}{5}{\cos }^{-1}\left(x^{5/2}\right)+c$
• C. $\frac{2}{5}{\cosh }^{-1}\left(x^{5/2}\right)+c$
• D. $\frac{2}{3}{\cos }^{-1}\left(x^{5/2}\right)+c$

Answer 1

The correct option is C $\frac{2}{5}{\cosh }^{-1}\left(x^{5/2}\right)+c$

$\int \sqrt{\frac{x^3}{x^5-1}}dx=\int \frac{x^{\frac{3}{2}}}{\sqrt{(x^{\frac{5}{3}}{)}^2}-1}dx$
$x^{\frac{5}{2}}=t$
$\frac{5}{2}{x}^{\frac{3}{2}}dx=dt{x}^{\frac{3}{2}}dx=\frac{2}{5}dt$
$\int \frac{\frac{2}{5}dt}{\sqrt{t^2-1}}=\frac{2}{5}\int \frac{dt}{\sqrt{t^2-1}}$
$\frac{2}{5}\log \mid t+\sqrt{t^2-1}\mid +c$
$=\frac{2}{5}\cos h^{-1}t+c$
$=\frac{2}{5}\cos h^{-1}(x^{\frac{5}{2}}+c$