# Solve \int \frac{e^{x}+1}{e^{x}(e^{2x}-1)}dx in two ways. Use a few substitutions the first time...

## Question:

Solve {eq}\int \frac{e^{x}+1}{e^{x}(e^{2x}-1)}dx {/eq} in two ways. Use a few substitutions the first time around - no partial fractions. Show me these substitutions. Then do the problem again using partial fractions. show me the set-up and state the final set of coefficients - no arithmetic needed to be seen. Your efforts will help illustrate to you, how different techniques give rise to different kinds of challenges & simplifications. Summarize your work and put the solution in the box.

## Integration:

Integration is a process to find the area under the curve and to find the volume.

It is a reverse process of differentiation.

Integration can be done using the substitution method or by partial fractions.

We know that {eq}\int \frac{dx}{x}=\ln x\,,\,\int x^n\,dx=\frac{x^{n+1}}{n+1} {/eq}

By substitution:

{eq}\int \frac{e^{x}+1}{e^{x}(e^{2x}-1)}dx\\ = \int \frac{e^{-x}\left ( e^{x}+1 \right )}{(e^{2x}-1)}dx {/eq}

Take {eq}e^{-x}=u\Rightarrow -e^{-x}\,dx=du {/eq}

{eq}\int \frac{e^{x}+1}{e^{x}(e^{2x}-1)}dx\\ = \int \frac{e^{-x}\left ( e^{x}+1 \right )}{(e^{2x}-1)}dx\\ =\int \frac{u}{u-1}\,du\\ =\int \frac{u-1+1}{u-1}\,du\\ =\int 1+\frac{1}{u-1}\,du\\ =u+\ln (u-1)+C\,\,\left \{ \because \int \frac{du}{u}=\ln u\,,\,\int du=u \right \} {/eq}

Put {eq}u=e^{-x} {/eq}

Therefore,

{eq}\int \frac{e^{x}+1}{e^{x}(e^{2x}-1)}dx=e^{-x}+\ln (e^{-x}-1)+C {/eq}

By Partial Fractions:

{eq}\int \frac{e^{x}+1}{e^{x}(e^{2x}-1)}dx= \int \frac{e^{x}+1}{e^{x}(e^{x}-1)(e^{x}+1)}dx\\ = \int \frac{dx}{e^{x}(e^{x}-1)} {/eq}

Let {eq}e^x=t\Rightarrow e^x\,dx=dt {/eq}

Therefore,

{eq}\int \frac{e^{x}+1}{e^{x}(e^{2x}-1)}dx= \int \frac{e^{x}+1}{e^{x}(e^{x}-1)(e^{x}+1)}dx\\ = \int \frac{dx}{e^{x}(e^{x}-1)}\\ =\int \frac{dt}{t^2(t-1)}\\ =\int \frac{-1}{t}-\frac{1}{t^2}+\frac{1}{t-1}\,dt\\ =-\ln t+\frac{1}{t}+\ln (t-1)+C\,\,\left \{ \because \int \frac{dt}{t}=\ln t\,,\,\int t^n\,dt=\frac{t^{n+1}}{n+1} \right \} {/eq}

Put {eq}e^x=t {/eq}

So,

{eq}\int \frac{e^{x}+1}{e^{x}(e^{2x}-1)}dx= \int \frac{e^{x}+1}{e^{x}(e^{x}-1)(e^{x}+1)}dx\\ =-\ln t+\frac{1}{t}+\ln (t-1)+C\\ =-\ln e^x+\frac{1}{e^x}+\ln (e^x-1)+C {/eq} 