Can you integrate this?

We use the substitution $u = e^x \rightarrow \mathrm{d}u = e^x \, \mathrm{d}x$ . We have

$\begin{aligned} \displaystyle\int_0^{\ln 13} \dfrac {e^x (5 + e^x)}{\sqrt{e^{2x}+10e^x+1}} \, \mathrm{d}x &= \displaystyle\int_0^{13} \dfrac {5+u}{\sqrt{u^2+10u+1}} \, \mathrm{d}u. \end{aligned}$

Now, we use the substitution $v = u^2 + 10u + 1 \rightarrow \mathrm{d}v = (2u + 10) \, \mathrm{d}u$ . Our integral is now easy to evaluate: $\begin{aligned} \displaystyle\int_{12}^{300} \dfrac {\frac {1}{2} \, \mathrm{d}v}{\sqrt{v}} &= \displaystyle\int_{12}^{300} \dfrac {1}{2} v^{-\tfrac{1}{2}} \, \mathrm{d}v \\&= \left. \sqrt{v} \right|_{12}^{300} \\&= \sqrt{300} - \sqrt{12} \\&= 10\sqrt{3} - 2\sqrt{3} \\&= \boxed {8\sqrt{3} \approx 13.856}. \end{aligned}$

Note that $\displaystyle e^{2x} + 10e^x + 1 = e^x(5 + e^x) + 5e^x + 1$ . We make the substitution $\displaystyle u = 5 + e^x$ . Then $\displaystyle \frac{dx}{du} = e^{-x}$ and our integral becomes

$\displaystyle \int_6^{18} \frac{u}{\sqrt{u(u-5) + 5(u-5) + 1}} \ du$

$\displaystyle = \int_6^{18} \frac{u}{\sqrt{u^2 - 24}} \ du$

$\displaystyle = \left[\sqrt{u^2 - 24} \right]_6^{18} = \sqrt{300} - \sqrt{12} = 4 \sqrt{12}$

$\displaystyle = 8 \sqrt{3}$ .

$\begin{aligned} I & = \int_0^{\ln 13} \frac {e^x(5+e^x)}{\sqrt{e^{2x}+10e^x+1}} dx & \small \color{#3D99F6} \text{Let }u = e^x \implies du = e^x \ dx \\ & = \int_1^{13} \frac {5+u}{\sqrt{u^2+10u+1}} dx & \small \color{#3D99F6} \text{Let }v^2 = u^2+10u+1 \implies 2v \ dv = 2u + 10 \ du \\ & = \int_{\sqrt{12}}^{\sqrt{300}} \frac vv \ dv \\ & = \int_{2 \sqrt 3}^{10\sqrt 3} dv \\ & = 10\sqrt 3 - 2\sqrt 3 \\ & = 8 \sqrt 3 \\ & \approx \boxed{13.856} \end{aligned}$