Question

Solve : $\int \frac{1}{\sqrt{1-2^{ex}}}dx$

Answer 1

$I=\int \frac{dx}{\sqrt{1-2^{ex}}}$

$kt1-2^{ex}=p$

differentiate both side

$-2^{ex}.e.\log 2dx=dp$

$\Rightarrow dx=\frac{-dp}{\left(e\log 2\right)\left(2^{ex}\right)}$

$dx=\frac{-dp}{e\log 2(1-p)}$

$I=\int \frac{dp}{\sqrt{P}.e.\log 2(p-1)}$

$\Rightarrow =\frac{1}{e\log 2}\int \frac{dp}{\sqrt{P}(p-1)}$

$\Rightarrow =\frac{1}{e\log 2}\int \frac{dp}{P^{2/2}-p^{1/2}}$

let $P=t^2$

$dp=2tdt$

$\Rightarrow \frac{1}{e\log 2}\int \frac{2tdt}{(t^3-t)}$

$\Rightarrow \frac{1}{e\log 2}\int \frac{2tdt}{(t^2-1)}$

$\Rightarrow \frac{2}{e\log 2}\int \frac{dt}{(t+1)(t+1)}$

$\Rightarrow \frac{2}{e\log 2}[\int \frac{1}{2}\frac{1}{t-1}-\frac{1}{t+1}]dt$

$\Rightarrow \frac{1}{e\log 2}[\int \frac{1}{(t-1)}dt-\int \frac{dt}{(t+1)}]$

$\Rightarrow \frac{1}{e\log 2}\left[\log \right(t-1)-\log (t+1\left)\right]$

$\Rightarrow \frac{1}{e\log 2}\log \left|\frac{t-1}{t+1}\right|$

$\Rightarrow \frac{1}{e\log 2}\log \left|\frac{\sqrt{p}-1}{\sqrt{p}+1}\right|$

$\Rightarrow \frac{1}{e\log 2}\log \left|\frac{\sqrt{1-2^{ex}}-1}{\sqrt{1-2^{ex}}+1}\right|$