Question

∫(1/(√(x+√x)))dx

Answer 1

let u = sqrt(x)

\[\large \int\limits{dx \over \sqrt{x + \sqrt{x}}} = \int\limits {2u.du \over \sqrt{u^2 + u}}\]

\[\large = \int\limits {2\sqrt{u}.du \over \sqrt{u + 1}} = \int\limits {2\sqrt{u}.du \over \sqrt{(\sqrt{u})^2 + 1}}\]

let\[\large \sqrt{u} = \sinh{x}\]
\[\large \int\limits{2\sqrt{u}.du \over \sqrt{(\sqrt{u})^2 + 1}} = \int\limits{4\sinh^2(y)\cosh(y).dy \over \cosh(y)}\]

\[\large = \int\limits {4\sinh^2(y).dy }\]

\[\large \int\limits{2\cosh(2y)}.dy - 2\int\limits{dy}\]

Answer 2

\[\int\limits_{}^{}\frac{ \sqrt{x+\sqrt{x}} }{ \sqrt{x}\times(\sqrt{x}+1) }\]
\[\sqrt{x}=y \rightarrow dx=2\sqrt{x}\times y dy\]
\[2 \int\limits_{}^{} \sqrt{\frac{ y }{ y+1 }} = 2 \int\limits_{}^{} \sqrt{1-\frac{ 1 }{ y+1 }} \]
let y+1= \[\csc ^{2}u\]
then simplify and integrate by parts, that will get you an answer ....

Answer 3

\[\frac{ 1 }{ \sqrt{x} +\sqrt{x}}=\frac{ 1 }{ 2\sqrt{x} }\]

\[\int\limits_{}^{}\frac{ 1 }{ 2\sqrt{x} }dx=\sqrt{x}+c\]

Answer 4

@NoelGreco it's so nice to consider it this way lol