Question

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

Answer 1

this is answer ...........

Answer 2

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

Let $\bf{e^{2x}+e^{-2x}=f(x)}$

differentiate both sides,

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

now, put it above integration (I),

I=$\bf{\int{\frac{f'(x)}{2f(x)}}\,dx}$

= $\bf{\frac{1}{2}ln|f(x)|+C}$

put f(x) = $(e^{2x}+e^{-2x})$
therefore, I =$\bf{\frac{1}{2}ln|e^{2x}+e^{-2x}|+C}$