Question

$\int x^2\sqrt{x^6-1}dx=$

• A. $\frac{1}{6}\{x^3\sqrt{x^6-1}-\log (x^3+\sqrt{x^6-1})\}+c$
• B. $\frac{1}{6}\{x^3\sqrt{x^6-1}+\log (x^3+\sqrt{x^6-1})\}+c$
• C. $\frac{1}{6}\{x^3\sqrt{x^6-1}-{\sin }^{-1}\left(x^3\right)\}+c$
• D. $\frac{1}{6}\{x^3\sqrt{x^6-1}+{\sin }^{-1}\left(x^3\right)\}+c$

Answer 1

The correct option is A $\frac{1}{6}\{x^3\sqrt{x^6-1}-\log (x^3+\sqrt{x^6-1})\}+c$

$\int x^2\sqrt{x^6-1}dx$

$Applyu-substitution:u=x^6$

$=\int \frac{\sqrt{u-1}}{6\sqrt{u}}du$

$=\frac{1}{6}\cdot \int \frac{\sqrt{u-1}}{\sqrt{u}}du$

$ApplyIntegrationByParts:u=\sqrt{u-1},v^{\prime }=\frac{1}{\sqrt{u}}$

$=\frac{1}{6}(2u^{\frac{1}{2}}\sqrt{u-1}-\int \sqrt{\frac{u}{u-1}}du)$

$=\frac{1}{6}(2u^{\frac{1}{2}}\sqrt{u-1}-(\sqrt{\frac{u}{u-1}}(u-1)+\ln |\sqrt{u}+\sqrt{u-1}|))$

$=\frac{1}{6}(2\sqrt{x^6}\sqrt{x^6-1}-(\sqrt{\frac{x^6}{x^6-1}}(x^6-1)+\ln |\sqrt{x^6}+\sqrt{x^6-1}|))$

$=\frac{1}{6}(x^3\sqrt{x^6-1}-\ln |x^3+\sqrt{x^6-1}|)+C$