# In the given equation as follows , use a table

In the given equation as follows , use a table of integrals with forms involving the trigonometric functions to find the indefinite integral:
$\int \frac{1}{1+{e}^{2x}}dx$
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Bertha Jordan

Step 1
Let $1+{e}^{2x}=t$. Differentiate both sides.
${e}^{2x}×2dx=dt$
$dx=\frac{dt}{2{e}^{2x}}$
$=\frac{dt}{2\left(t-1\right)}$
Substitute the values into the integral. Also since the integral is indefinite; a constant of integration is to be added.
Step 2
Perform the integration.
$\int \frac{1}{1+{e}^{2x}}dx=\int \frac{1}{t}×\frac{dt}{2\left(t-1\right)}$
$=\frac{1}{2}\int \frac{dt}{t\left(t-1\right)}$
$=\frac{1}{2}\int \left[\frac{1}{t-1}-\frac{1}{t}\right]dt$
$=\frac{1}{2}×\left[\mathrm{ln}|t-1|-\mathrm{ln}|t|\right]+c$
$=\frac{1}{2}×\left[\mathrm{ln}|1+{e}^{2x}-1|-\mathrm{ln}|1+{e}^{2x}|\right]+c$
$=\frac{1}{2}×\left[\mathrm{ln}|{e}^{2x}|-\mathrm{ln}|1+{e}^{2x}|\right]+c$
Hence the solution is obtained.

###### Not exactly what you’re looking for?
SlabydouluS62
$\int \frac{1}{{e}^{2x}+1}dx$
$=-\frac{1}{2}\int \frac{1}{{e}^{-u}+1}du$
$=\int \frac{{e}^{u}}{{e}^{u}+1}du$
$=\int \frac{1}{v}dv$
$=\mathrm{ln}\left(v\right)$
$=\mathrm{ln}\left({e}^{u}+1\right)$
$-\frac{1}{2}\int \frac{1}{{e}^{-u}+1}du$
$=-\frac{\mathrm{ln}\left({e}^{u}+1\right)}{2}$
$=-\frac{\mathrm{ln}\left({e}^{u}+1\right)}{2}+C$