Question

Solve $I={\int }_0^8\frac{\sqrt{x}}{\sqrt{x}+\sqrt{8-x}}dx$

Answer 1

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

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

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

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

Put $x–4=u=⟹dx=du$ and $x^2=(u+4{)}^2$

$=\frac{1}{2}\frac{u+4-\sqrt{8u+32–(u+4{)}^2}}{u}du$

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

$v=u^2⟹du=\frac{1}{2u}dv$ and $\frac{1}{u^2}=\frac{1}{v}$

$=\frac{u}{2}+2logu-\frac{1}{4}\int \frac{\sqrt{16-v}}{v}dv$

$=fracu2+2logu-\frac{1}{2}\int \frac{w^2}{w^2-16}dw$ put $\sqrt{16–v}=w$ and $dv=-2\sqrt{16}-vdw$

$=\frac{u}{2}+2logu-\frac{1}{2}\int \frac{w^2+16-16}{w^2-16}dw$

$=\frac{u}{2}+2logu-\frac{1}{2}w–8\int \frac{dw}{w^2-16}$

$=\frac{u}{2}+2logu-\frac{1}{2}\sqrt{16–u^2}–8\times \frac{1}{2}log|\frac{w-4}{w+4}|$

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

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

$=\frac{8-4}{2}+2log|8-4|-\frac{1}{2}\sqrt{8.8–(8{)}^2}–log|\frac{\sqrt{8.8–8^2}-4}{\sqrt{8.8-8^2}+4}|$

$-(-2+2log4-0)=4$