 bobbie71G

2021-10-20

Evaluate the following integrals.
$\int \frac{{e}^{x}}{{e}^{2x}+2{e}^{x}+17}dx$ jlo2niT

Given that:
The integral is $\int \frac{{e}^{x}}{{e}^{2x}+2{e}^{x}+17}dx$
By using the formula,
$\int \frac{1}{{x}^{2}+{a}^{2}}dx=\frac{1}{a}\mathrm{arctan}\left(\frac{x}{a}\right)+C$
By using substitution,
Substitute $v={e}^{x}$, then $dv={e}^{x}dx$
Then,
$\int \frac{{e}^{x}}{{e}^{2x}+2{e}^{x}+17}dx=\int \frac{1}{{v}^{2}+2v+17}dv$
$=\int \frac{1}{{v}^{2}+2v+1-1+17}dv$
$=\int \frac{1}{\left({v}^{2}+2v+1\right)+16}dv$
$=\int \frac{1}{{\left(v+1\right)}^{2}+16}dv$
Substitute u=v+1, du=dv
Then, $=\int \frac{1}{{u}^{2}+16}du$
$=\int \frac{1}{{u}^{2}+{4}^{2}}du$
$=\frac{1}{4}\mathrm{arctan}\left(\frac{u}{4}\right)$
Substitute back $u=v+1$ and $v={e}^{x}$, then $u={e}^{x}+1$
$\int \frac{{e}^{x}}{{e}^{2x}+2{e}^{x}+17}dx=\frac{1}{4}\mathrm{arctan}\left(\frac{u}{4}\right)$
$=\frac{1}{4}\mathrm{arctan}\left(\frac{{e}^{x}+1}{4}\right)+C$
Where C is the integration constant.
Therefore,
$\int \frac{{e}^{x}}{{e}^{2x}+2{e}^{x}+17}dx=\frac{1}{4}\mathrm{arctan}\left(\frac{{e}^{x}+1}{4}\right)+C$, where c is the integration constant.

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