Question

Q.63. $\int \frac{x^5}{\sqrt{1+x^3}}dx$

Answer 1

Dear student

$$\begin{gathered} \int \frac{{\mathrm{x}}^5}{\sqrt{1+{\mathrm{x}}^3}}\mathrm{dx} \\ \mathrm{Put}\mathrm{u}=1+{\mathrm{x}}^3\Rightarrow \mathrm{dx}=\frac{1}{3{\mathrm{x}}^2}\mathrm{du} \\ =\frac{1}{3}\int \frac{\mathrm{u}-1}{\sqrt{\mathrm{u}}}\mathrm{du} \\ =\frac{1}{3}\int \left(\sqrt{\mathrm{u}}-\frac{1}{\sqrt{\mathrm{u}}}\right)\mathrm{du} \\ =\frac{{\displaystyle 1}}{{\displaystyle 3}}\int \sqrt{\mathrm{u}}\mathrm{du}-\frac{{\displaystyle 1}}{{\displaystyle 3}}\int \frac{{\displaystyle 1}}{{\displaystyle \sqrt{\mathrm{u}}}}\mathrm{du} \\ =\frac{1}{3}\left(\frac{{\mathrm{u}}^{{\displaystyle \frac{1}{2}}+1}}{\frac{1}{2}+1}\right)-\frac{1}{3}\left(\frac{{\mathrm{u}}^{{\displaystyle \frac{-1}{2}}+1}}{\frac{-1}{2}+1}\right)+\mathrm{C} \\ =\frac{2{\mathrm{u}}^{{\displaystyle \frac{3}{2}}}}{9}-\frac{2\sqrt{\mathrm{u}}}{3}+\mathrm{C} \\ =\frac{{\displaystyle 2{\left(1+{\mathrm{x}}^3\right)}^{\frac{3}{2}}}}{9}-\frac{2\sqrt{1+{\mathrm{x}}^3}}{3}+\mathrm{C} \\ =\frac{2{\left(1+{\mathrm{x}}^3\right)}^{\frac{3}{2}}-6\sqrt{1+{\mathrm{x}}^3}}{9}+\mathrm{C} \\ =\frac{{\displaystyle 2\sqrt{1+{\mathrm{x}}^3}\left(1+{\mathrm{x}}^3\right)-6\sqrt{1+{\mathrm{x}}^3}}}{{\displaystyle 9}}+\mathrm{C} \\ =\frac{2\sqrt{1+{\mathrm{x}}^3}\left({\mathrm{x}}^3-2\right)}{9}+\mathrm{C} \end{gathered}$$
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