Question

Evaluate the integration by substitution ∫▒〖1/√(1+√(1+x)) dx〗

Answer 1

Given integral is-

$\int\dfrac{1}{\sqrt{1+\sqrt{1+x}}}dx$ \

Let

$1+x=t^2\\ \Rightarrow dx=2tdt$

$\implies Integral =\int\dfrac{2t}{\sqrt{1+t}}dt$

$=\int 2\dfrac{(1+t-1)}{\sqrt{1+t}}dt$

$=\int 2({\sqrt{1+t}}-\dfrac{1}{\sqrt{1+t}})dt$

$=2(\dfrac{2(1+t)^{\dfrac{3}{2}}}{3}-2\sqrt{1+t})+C$

$=\dfrac{4}{3}((1+\sqrt{1+x})^{\dfrac{3}{2}})-4\sqrt{1+\sqrt{1+x}}+C$