Question

$\underset{0}{\overset{5}{\int }}\frac{\sqrt[4]{x+4}}{\sqrt[4]{x+4}+\sqrt[4]{9-x}}dx$

Answer 1

$$\begin{gathered} \mathrm{Let}I={\int }_0^5\frac{\sqrt[4]{x+4}}{\sqrt[4]{x+4}-\sqrt[4]{9-x}}dx...\left(\mathrm{i}\right) \\ I={\int }_0^5\frac{\sqrt[4]{9-x}}{\sqrt[4]{9-x}-\sqrt[4]{x+4}}dx \\ \left[\mathrm{Using}{\int }_0^{\mathrm{a}}\mathrm{f}\left(\mathrm{x}\right)\mathrm{d}\mathrm{x}={\int }_0^{\mathrm{a}}\mathrm{f}\left(\mathrm{a}-\mathrm{x}\right)\mathrm{d}\mathrm{x}\right] \\ \mathrm{I}=-{\int }_0^5\frac{\sqrt[4]{9-\mathrm{x}}}{\sqrt[4]{\mathrm{x}+4}-\sqrt[4]{9-\mathrm{x}}}\mathrm{d}\mathrm{x}...\left(\mathrm{ii}\right) \\ \mathrm{Adding}\left(\mathrm{i}\right)\mathrm{and}\left(\mathrm{ii}\right) \\ 2I={\int }_0^5\frac{\sqrt[4]{x+4}}{\sqrt[4]{x+4}-\sqrt[4]{9-x}}-\frac{\sqrt[4]{9-x}}{\sqrt[4]{x+4}-\sqrt[4]{9-x}}dx \\ ={\int }_0^5\frac{\sqrt[4]{x+4}-\sqrt[4]{9-x}}{\sqrt[4]{x+4}-\sqrt[4]{9-x}}dx \\ ={\int }_0^{5}dx \\ ={\left[x\right]}_0^5 \\ =5 \\ \mathrm{Hence}I=\frac{5}{2} \end{gathered}$$