MHB How to Integrate 4e^x/(16e^2x+25)?

[karush]

$\large{242.7.5.89}$
answer by Maxima
$$\displaystyle
I_{89}=\int\frac{4e^{x}}{16e^{2x}+25}\, dx
= \dfrac{\arctan\left(\frac{4\mathrm{e}^x}{5}\right)}{5}+C
\\
\begin{align}
\displaystyle
u& = { {4x}^{3}} & \frac{1}{u} du&= \,dx
\end{align} \\$$
so by table..
$$\displaystyle
I_{89}=\int\frac{u}{u^2+5^2}\frac{1}{u} du
=\int\frac{1\, }{u^2+5^2} du
=\frac{\tan^{-1}\left({\frac{u}{5}}\right)}{5}+C$$so far?

 

[MarkFL]

I think I would write:

$$I_{89}=\int\frac{4e^x}{\left(4e^x\right)^2+25}\,dx$$

Let:

$$u=4e^x\implies du=4e^x\,dx\implies dx=\frac{du}{u}$$

And now you have:

$$I_{89}=\int\frac{1}{u^2+5^2}\,du$$

 

[karush]

.
$$\displaystyle
I_{89}=\int\frac{u}{u^2+5^2}\cdot\frac{du}{u}
=\int\frac{1 }{u^2+5^2} du
=\frac{\tan^{-1}\left({\frac{u}{5}}\right)}{5}+C$$

back substitute $u=4e^{x}$

$$\frac{\tan^{-1}\left({\frac{4e^{x}}{5}}\right)}{5}+C$$

kinda maybe??

 

[MarkFL]

Looks good to me! (Yes)

 

[karush]

think I am getting it
30 in the class Friday... nice teacher.