How Can We Integrate This Complex Logarithmic Function?

[karush]

$\tiny\text{LCC 206 {7.R.31} Integral log }$
$$\displaystyle
I=\int_0^{\ln\left({10}\right)}
\frac{e^{x}\sqrt{e^{x}-1}}{e^{x}+8}
\ dx \\

\begin{align}\displaystyle
u& = e^{x}-1 &
du&= e^{x} \ d{x} \\
\end{align} \\

I=\int_0^{\ln\left({10}\right)}
\frac{\sqrt{u}}{u+9}
\ du $$
Couldn't get this to integrate

$\tiny\text
{from Surf the Nations math study group}$

 

[Prove It]

$\tiny\text{LCC 206 {7.R.31} Integral log }$
$$\displaystyle
I=\int_0^{\ln\left({10}\right)}
\frac{e^{x}\sqrt{e^{x}-1}}{e^{x}+8}
\ dx \\

\begin{align}\displaystyle
u& = e^{x}-1 &
du&= e^{x} \ d{x} \\
\end{align} \\

I=\int_0^{\ln\left({10}\right)}
\frac{\sqrt{u}}{u+9}
\ du $$
Couldn't get this to integrate

$\tiny\text
{from Surf the Nations math study group}$

I would have gone with the substitution $\displaystyle \begin{align*} u = \mathrm{e}^x + 8 \implies \mathrm{d}u = \mathrm{e}^x\,\mathrm{d}x \end{align*}$, notice that $\displaystyle \begin{align*} u(0) = 9 \end{align*}$ and $\displaystyle \begin{align*} u\,\left( \ln{(10)} \right) = 18 \end{align*}$ and the integral becomes

$\displaystyle \begin{align*} \int_0^{\ln{(10)}}{\frac{\mathrm{e}^x\,\sqrt{ \mathrm{e}^x - 1 }}{\mathrm{e}^x + 8} \,\mathrm{d}x} &= \int_9^{18}{ \frac{\sqrt{u - 9}}{u}\,\mathrm{d}u } \\ &= 2\int_9^{18}{ \frac{u - 9}{2\,u\,\sqrt{u - 9}}\,\mathrm{d}u } \end{align*}$

Now let $\displaystyle \begin{align*} v = \sqrt{u - 9} \implies \mathrm{d}v = \frac{1}{2\,\sqrt{u - 9}}\,\mathrm{d}u \end{align*}$ and note that $\displaystyle \begin{align*} v(9) = 0 \end{align*}$ and $\displaystyle \begin{align*} v(18) = 3 \end{align*}$ and the integral becomes

$\displaystyle \begin{align*} 2\int_9^{18}{\frac{u-9}{2\,u\,\sqrt{u - 9}}\,\mathrm{d}u} &= 2 \int_0^3{ \frac{v^2}{ v^2 + 9 } \, \mathrm{d}v } \end{align*}$

Now let $\displaystyle \begin{align*} v = 3\tan{ \left( \theta \right) } \implies \mathrm{d}v = 3\sec^2{ \left( \theta \right) } \,\mathrm{d}\theta \end{align*}$ and note that when $\displaystyle \begin{align*} v = 0 , \, \theta = 0 \end{align*}$ and when $\displaystyle \begin{align*} v = 3, \, \theta = \frac{\pi}{4} \end{align*}$ and the integral becomes

$\displaystyle \begin{align*} 2\int_0^3{ \frac{v^2}{v^2 + 9}\,\mathrm{d}v } &= 2\int_0^{\frac{\pi}{4}}{ \frac{\left[ 3\tan{ \left( \theta \right) } \right] ^2 }{ \left[ 3\tan{ \left( \theta \right) } \right] ^2 + 9 }\, 3\sec^2{ \left( \theta \right) } \,\mathrm{d}\theta } \\ &= 6\int_0^{\frac{\pi}{4}}{ \frac{9\tan^2{ \left( \theta \right) }\sec^2{\left( \theta \right) } }{ 9\sec^2{ \left( \theta \right) } }\, \mathrm{d}\theta } \\ &= 6\int_0^{\frac{\pi}{4}}{ \tan^2{ \left( \theta \right) } \,\mathrm{d}\theta } \\ &= 6 \int_0^{\frac{\pi}{4}}{ \left[ \sec^2{ \left( \theta \right) } - 1 \right] \,\mathrm{d}\theta } \\ &= 6 \, \left[ \tan{ \left( \theta \right) } - \theta \right] _0^{\frac{\pi}{4}} \\ &= 6\,\left\{ \left[ \tan{ \left( \frac{\pi}{4} \right) } - \frac{\pi}{4} \right] - \left[ \tan{ \left( 0 \right) } - 0 \right] \right\} \\ &= 6 \,\left( 1 - \frac{\pi}{4} \right) \\ &= \frac{6\,\left( 4 - \pi \right) }{4} \\ &= \frac{3\,\left( 4 - \pi \right) }{2} \end{align*}$

 

[karush]

$\tiny\text{LCC 206 {7.R.31} Integral log }$

OK I continued with my original but yours was easier
$$\displaystyle
I=\int_0^{\ln\left({10}\right)}
\frac{e^{x}\sqrt{e^{x}-1}}{e^{x}+8}
\ dx \\

\begin{align}\displaystyle
u& = e^{x}-1 &
du&= e^{x} \ d{x} \\
\end{align}
\\
I=\int_0^{9}
\frac{\sqrt{u}}{u+9} \ du $$
\begin{align}\displaystyle
u& = 9\tan^2 \left({w}\right) &
du&= \frac{18\sin\left({w}\right)}
{\cos^3 \left({w}\right)} \ d{w} \\
\end{align}

$6\int_0^{\pi/4}\tan^2 \left({w}\right) \ dw =\left[-6-\tan\left({w}\right)\right]_0^{\pi /4 }
6-\frac{3 \pi}{2}+C$ $\tiny\text
{from Surf the Nations math study group}$