Question

If $I=\int \frac{x^8}{\sqrt{1+x^3}}dx$, then $I$ equals

• A. $\frac{2}{3}\sqrt{1+x^3}\left[\frac{1}{5}\right(1+x^3{)}^2+\frac{2}{3}(1+x^3)+1]+C$
• B. $\frac{2}{3}\sqrt{1+x^3}[\frac{1}{5}{x}^6+\frac{4}{15}{x}^3+\frac{8}{15}]+C$
• C. $\frac{2}{45}\sqrt{1+x^3}(3x^6-4x^3+8)+C$
• D. None of these

Answer 1

Answer: (C)

$\frac{2}{45}\sqrt{1+x^3}(3x^6-4x^3+8)+C$

$I=\int \frac{x^8}{\sqrt{1+x^3}}dx$
Substitute $t=x^3\Rightarrow dt=3x^{2}dx$
$\therefore I=\frac{1}{3}\int \frac{t^2}{\sqrt{t+1}}dt$
Substitute $u=t+1\Rightarrow du=dt$
$\therefore I=\frac{1}{3}\int \frac{{(u-1)}^2}{\sqrt{u}}du$
$=\frac{1}{3}\int (u^{\frac{3}{2}}-2\sqrt{u}+\frac{1}{\sqrt{u}})du$
$=\frac{2u^{\frac{5}{2}}}{15}-\frac{4u^{\frac{3}{2}}}{9}+\frac{2\sqrt{u}}{3}+C$
$=\frac{2}{45}\sqrt{x^3+1}(3x^6-4x^3+8)+C$