Question

Solve the following : ${\displaystyle {\int }_0^1\frac{1}{\sqrt{1+x}+\sqrt{x}}dx}$

Answer 1

Let I = ${\displaystyle {\int }_0^1\frac{1}{\sqrt{1+x}+\sqrt{x}}dx}$

= ${\displaystyle {\int }_0^1\frac{1}{\sqrt{1+x}+\sqrt{x}}\times \frac{\sqrt{1+x}-\sqrt{x}}{\sqrt{1+x}-\sqrt{x}}dx}$

= ${\displaystyle {\int }_0^1\frac{\sqrt{1+x}-\sqrt{x}}{{\left(\sqrt{1+x}\right)}^2-\left({\sqrt{x}}^2\right)}}$dx

= ${\displaystyle {\int }_0^1\frac{\sqrt{1+x}-\sqrt{x}}{1+x-x}dx}$

= ${\displaystyle {\int }_0^1[{(1+x)}^{\frac{1}{2}}-x^{\frac{1}{2}}]dx}$

= ${\displaystyle {\int }_0^1{(1+x)}^{\frac{1}{2}}dx-{\int }_0^1{x}^{\frac{1}{2}}dx}$

= ${\displaystyle {\left[\frac{{(1+x)}^{\frac{1}{2}}}{\frac{3}{2}}\right]}_0^1-{\left[\frac{x^{\frac{3}{2}}}{\frac{3}{2}}\right]}_0^1}$

= ${\displaystyle \frac{2}{3}[{\left(2\right)}^{\frac{3}{2}}-{\left(1\right)}^{\frac{3}{2}}]-\frac{2}{3}[{\left(1\right)}^{\frac{3}{2}}-0]}$

= ${\displaystyle \frac{2}{3}(2\sqrt{2}-1)-\frac{2}{3}\left(1\right)}$

= ${\displaystyle \frac{4\sqrt{2}}{3}-\frac{2}{3}-\frac{2}{3}}$

∴ I = ${\displaystyle \frac{4}{3}(\sqrt{2}-1)}$