Question

Evaluate: ${\int }_0^1\frac{1}{\sqrt{1+x+\sqrt{x}}}dx$

Answer 1

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

$u+v=\sqrt{x}$

$\Rightarrow dx=2\sqrt{x}du$

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

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

For $\int \frac{2u+1}{2\sqrt{u^2+u+1}}du$

$u+v=u^2+v+1$

$\Rightarrow du=\frac{1}{2u+1}dv$

$=\int \frac{1}{\sqrt{v}}dv$

$=2\sqrt{v}$

$=2\sqrt{v^2+v+1}$

For $=\int \frac{1}{\sqrt{u^2+u+1}}du$

$=-\int \frac{1}{\sqrt{{(1+\frac{1}{2})}^2+\frac{3}{4}}}du$

$u+v=\frac{2v+1}{\sqrt{3}}$

$\Rightarrow du=\frac{\sqrt{3}}{2}dv$

$=\int \frac{1}{\sqrt{v^2+1}}dv$

$=ln(\sqrt{v^2+1}+v)$

$=ln(\sqrt{{\left(\frac{2v+1}{\sqrt{3}}\right)}^2+1}+\frac{2v+1}{\sqrt{3}})$

So integral is

$2\sqrt{x+\sqrt{x}+1}=ln(2\sqrt{x+\sqrt{x}+1}+\sqrt{x})+1)+c$

Putting the values

$=0.6964$.