Question

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

Answer 1

$$\begin{gathered} \int \frac{x^{5}dx}{\sqrt{1+x^3}} \\ =\int \frac{x^3.x^{2}dx}{\sqrt{1+x^3}} \\ \mathrm{Let}1+x^3=t \\ \Rightarrow x^3=t-1 \\ \Rightarrow 3x^2=\frac{dt}{dx} \\ \Rightarrow x^{2}dx=\frac{dt}{3} \\ \mathrm{Now},\int \frac{x^3.x^{2}dx}{\sqrt{1+x^3}} \\ =\frac{1}{3}\int \frac{\left(t-1\right)}{\sqrt{t}}dt \\ =\frac{1}{3}\int \left(\sqrt{t}-\frac{1}{\sqrt{t}}\right)dt \\ =\frac{1}{3}\int \left(t^{\frac{1}{2}}-t^{-\frac{1}{2}}\right)dt \\ =\frac{1}{3}\left[\frac{t^{{\displaystyle \frac{1}{2}}+1}}{{\displaystyle \frac{1}{2}}+1}-\frac{t^{-{\displaystyle \frac{1}{2}}+1}}{-{\displaystyle \frac{1}{2}}+1}\right]+C \\ =\frac{1}{3}\left[\frac{2}{3}{t}^{\frac{3}{2}}-2\sqrt{t}\right]+C \\ =\frac{2}{9}{\left(1+x^3\right)}^{\frac{3}{2}}-\frac{2}{3}{\left(1+x^3\right)}^{\frac{1}{2}}+C \end{gathered}$$