# int_0^ln3 e^(3x)/(e^(6x)+5)dx. Help please?

May 2, 2018

$I = \frac{\sqrt{5}}{15} \left(\arctan \left(\frac{27}{\sqrt{5}}\right) - \arctan \left(\frac{1}{\sqrt{5}}\right)\right)$
$I \approx 0.15915$

#### Explanation:

We wish to solve ${\int}_{0}^{\ln 3} \frac{{e}^{3 x}}{{e}^{6 x} + 5} \mathrm{dx}$.

We note that ${e}^{6 x} = {\left({e}^{3 x}\right)}^{2}$ and make the u-substitution $u = {e}^{3 x}$. It follows that $\mathrm{du} = 3 {e}^{3 x} \mathrm{dx}$.

We prepare our integral for the substitution by multiplying by $\frac{3}{3}$.

$\frac{1}{3} {\int}_{0}^{\ln 3} \frac{3 {e}^{3 x}}{{\left({e}^{3 x}\right)}^{2} + 5} \mathrm{dx}$

Note that our lower bound becomes ${e}^{3 \cdot 0} = 1$ and our upper bound becomes ${e}^{3 \ln 3} = {e}^{\ln \left({3}^{3}\right)} = {3}^{3} = 27$.

$\frac{1}{3} {\int}_{1}^{27} \frac{1}{{u}^{2} + 5} \mathrm{du}$

This nearly looks like the proper form to yield $\arctan \left(u\right)$, but the $5$ needs to be a $1$. We factor out the $5$ and make another substitution.

$\frac{1}{3} {\int}_{1}^{27} \frac{1}{5 \left({u}^{2} / 5 + 1\right)} \mathrm{du} = \frac{1}{15} {\int}_{1}^{27} \frac{1}{{\left(\frac{u}{\sqrt{5}}\right)}^{2} + 1} \mathrm{du}$

Let $v = \frac{u}{\sqrt{5}}$. Then $\mathrm{dv} = \frac{1}{\sqrt{5}} \mathrm{du}$. This yields the integral:

$\frac{\sqrt{5}}{15} {\int}_{\frac{1}{\sqrt{5}}}^{\frac{27}{\sqrt{5}}} \frac{1}{{v}^{2} + 1} \mathrm{dv}$

This is the proper form to yield $\arctan \left(v\right)$. Evaluating yields:

$\frac{\sqrt{5}}{15} {\left[\arctan v\right]}_{\frac{1}{\sqrt{5}}}^{\frac{27}{\sqrt{5}}}$
$= \frac{\sqrt{5}}{15} \left(\arctan \left(\frac{27}{\sqrt{5}}\right) - \arctan \left(\frac{1}{\sqrt{5}}\right)\right)$
$\approx 0.15915$