Question

integrate 2^x.dx/(1-4^x)^1/2​

Answer 1

$\underline{\boxed{\red{Answer}}}$

$\rm{\blue{GIVEN\:QUESTION\;IS}}$

$\displaystyle\int{\Big(\frac{2^x}{\sqrt{1-4^x}}\Big)dx}$

$\underline{\mathbb{\EXPLINATION}}$

$\rm{\red{put\:\:2^x=z}}$

$\rm{Now\:Differentiate\:Both\:Sides\:w.r.t.x\: We\:Have}$

$\rm{\frac{dz}{dx}=2^x\:\log_{e}\left(2\right)}$

$\rm{\frac{dz}{\Big(2^x\log_{e}\left(2\right)\Big)}=dx}$

$\implies{\displaystyle\int{\Big(\frac{z}{\sqrt{1-z^2}}\Big)dx}}$

$\implies{\displaystyle\int{\Big(\frac{z}{\sqrt{1-z^2}}\times\:\frac{dz}{z\log_{e}\left(2\right)}\Big)}}$

$\implies{\rm{\frac{1}{\log_{e}\left(2\right)}}}$ $\displaystyle\int{\Big(\frac{dz}{\sqrt{1-z^2}}\Big)}$

$\implies{\rm{\frac{1}{\log_{e}\left(2\right)}}}$ $\rm{\left(Sin^{-1}\left(z\right)\right)+c}$

$\implies{\rm{\frac{Sin^{-1}\left(2^x\right)}{\log_{e}\left(2\right)}}+c}$

$\underline{\boxed{\pink{Where\:c\:is\:Called\:Constant\:of\:Integration}}}$

Answer 2

Answer:

Hope it will help you ☺️