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

This is an integration by parts problem, but first, set $u=x^3$. Then $ du = 3x^2 dx $, so the integral becomes
$$ \frac{1}{3}\int \frac{u}{\sqrt{1+u}} \, du, $$
which you can see how to do parts on:
$$ \int \frac{u}{\sqrt{1+u}} \, du = 2u\sqrt{1+u} - 2\int \sqrt{1+u} \, du = 2u\sqrt{1+u} - \frac{4}{3}(1+u)^{3/2} +C, $$
which you can also write as $\frac{2}{3}(u-2)\sqrt{1+u} + C$. Then the final answer is
$$ \frac{2}{9}(x^3-2)\sqrt{1+x^3} + C. $$
