Integrate $\int \frac{x^5 dx}{\sqrt{1+x^3}}$ I took $1+x^3$ as $t^2$ . I also split $x^5$ as $x^2 .x^3$ . Then I subsituted the differentiated value in in $x^2$ . I put $x^3$ as $1- t^2$ . I am getting the last step as $2/9[\sqrt{1+x^3}x^3 ]$ but this is the wrong answer , i should get $2/9[\sqrt{1+x^3}(x^3 +2)]$.
Please help me. Thanks

Substitute $t=1+x^3$ and $dt=3x^2$:
\begin{align*}
\int \frac{x^5 dx}{\sqrt{1+x^3}}&=\frac{1}{3}\int \frac{(t-1) dt}{\sqrt{t}}\\
&=\frac{1}{3}\int \left( \sqrt{t}-\frac{1}{\sqrt{t}}\right)dt\\
&=\frac{1}{3}\left( \frac{2}{3}t^\frac{3}{2}-2\sqrt{t}\right)\\
&=\frac{2}{9}(x^3-2)\sqrt{1+x^2}
\end{align*}
And of course don't forget the constant of integration.
