Question

Solve $\underset{0}{\overset{x}{\int }}\frac{\sqrt{x}}{\sqrt{x}+\sqrt{8-x}}dx$

Answer 1

$\underset{0}{\overset{x}{\int }}\frac{\sqrt{x}}{\sqrt{x}+\sqrt{8-x}}dx$

${\int }_0^x\frac{\sqrt{x}}{\sqrt{x}+\sqrt{8-x}}\times \frac{\sqrt{x}-\sqrt{8-x}}{\sqrt{x}-\sqrt{8-x}}dx$

${\int }_0^x\frac{x-\sqrt{8x-x^2}}{2x-8}dx$

$\frac{1}{2}{\int }_0^x\frac{x-\sqrt{8x-x^2}}{x-4}dx$

$\frac{1}{2}\int \frac{u+4-\sqrt{8u+32-{\left(u+4\right)}^2}}{u}du$

$\frac{1}{2}\int du+\int \frac{du}{u}-\frac{1}{2}\int \frac{\sqrt{-{\left(u+4\right)}^2+8u+32}}{u}du$

$v=u^2=du=\frac{1}{2u}du\int \frac{1}{u^2}=\frac{1}{v}$

$\frac{4}{2}+2\log u-\frac{1}{4}\int \frac{\sqrt{16-v}}{v}du$

$\frac{4}{2}+2\log u-\frac{1}{2}\int \frac{w^2}{w^2-16}dw$

$\frac{4}{2}+2\log u-\frac{1}{2}\int \frac{w^2+16-16}{w^2-16}dw$

$\frac{4}{2}+2\log u-\frac{1}{2}w-8\int \frac{dw}{w^2-16}$

$\frac{4}{2}+2\log u-\frac{1}{2}\sqrt{16-u^2}-8\times \frac{1}{8}\log \left|\frac{w-4}{w+4}\right|$

$\frac{x-4}{2}+2\log \left|x-4\right|-\frac{1}{2}\sqrt{16-{\left(x-4\right)}^2}-\log \left|\frac{\sqrt{16-u^2}-4}{\sqrt{16-u^2}+4}\right|$

$\frac{x-4}{2}+2\log \left|x-4\right|-\frac{1}{2}\sqrt{8x-x^2}-\log {\left|\frac{\sqrt{8x-x^2}-4}{\sqrt{8x-x^2}+4}\right|}_0^x$

$\frac{x-4}{2}+2\log \left|x-4\right|-\frac{1}{2}\sqrt{8x-x^2}-\log \left|\frac{\sqrt{8x-x^2}-4}{\sqrt{8x-x^2}+4}\right|$