Q.$\dfrac{e^{x}-e^{-x}}{e^{x}+e^{-x}}$

The substitution (change-of-variable) method mirrors the chain rule of differentiation. If $u=g(x)$ is a differentiable function, then

because $du=g'(x)\,dx$. Choosing $u$ so that its derivative already appears (up to a constant) in the integrand converts a hard integral into a standard one; after integrating in $u$, substitute back $u=g(x)$.

Two especially useful consequences (with $u=f(x)$):

Standard log-form results that follow are $\int \tan x\,dx=\log|\sec x|+c$, $\int \cot x\,dx=\log|\sin x|+c$, $\int \operatorname{cosec} x\,dx=\log|\operatorname{cosec} x-\cot x|+c$, and $\int \sec x\,dx=\log|\sec x+\tan x|+c$.

For a trigonometric substitution (e.g. $x=a\tan\theta$), draw a right triangle to read back the other trig ratios when reversing the substitution.