Question

integrate the function [(e^{2x} - 1)/(e^{2x} + 1)].dx

Answer 1

$\bf{I=\int{\frac{e^{2x}-1}{e^{2x}+1}}\,dx}$

we know, $\frac{e^{2x}-1}{e^{2x}+1}=\frac{e^x-e^{-x}}{e^x-e^{-x}}$

$\bf{I=\int{\frac{e^x-e^{-x}}{e^x+e^{-x}}}\,dx}$

Let $e^x + e^{-x} = f(x)$ -----(1)

differentiate both sides,

$e^x - e^{-x}=f'(x)$ -----(2)

put equations (1) and (2) in I,

I = $\bf{\int{\frac{f'(x)}{f(x)}}\,dx}$

= $\bf{log|f(x)|}+C$

put f(x) = $e^x + e^{-x}$

therefore, I = $\bf{log|e^x+e^{-x}|}+C$

Answer 2

HELLO DEAR,

given function is $\bold{\int{\frac{e^{2x} - 1}{e^{2x} + 1}}\,dx}$

Divide both numerator and denominator by e^{x}

$\bold{\int{\frac{\frac{e^{2x} - 1}{e^x}}{\frac{e^{2x} + 1}{e^x}}}\,dx}$

$\bold{\int{\frac{(e^x - e^{-x})}{(e^x + e^{-x})}}\,dx}$

let (e^x + e^{-x}) = t
$\bold{dt/dx = (e^x - e^{-x})}$
$\bold{\Rightarrow dx = dt/(e^x - e^{-x})}$

So, $\bold{\int{\frac{(e^x - e^{-x})}{(e^x + e^{-x})} * dt/(e^x - e^{-x})}\,dx}$

$\bold{\int{dt/t}}$

$\bold{log|t| + c}$

put the value of t in the above function,

$\bold{log|e^x + e^{-x}| + c}$


I HOPE ITS HELP YOU DEAR,
THANKS