1. ## U substitution question

Let u = e6x + 9. Then du = 6e6x dx and e6x dx =
 1 6
du, so

 e6x e6x + 9
dx
=
 1 u
 1 6
du
=
 1 6
ln|u| + C =
 1 6
ln(e6x + 9) + C.

Is this correct?

edit: my syntax is messed up sorry I will fix

2. ## Re: U substitution question

∫▒〖e^6x/(e^6x+9) dx〗 let e^(6x )=u Therefore 6e^6x dx=du
hence
∫▒〖e^6x/(e^6x+9) dx〗 = 1/6 ∫▒〖1/(u+9) du〗 Now you can finish it

3. ## Re: U substitution question

$\displaystyle \int \frac{e^{6x}}{e^{6x}+9}\,\, dx$

Let $\displaystyle u = e^{6x} + 9\,\,\therefore du = 6e^{6x}\,\,dx$

Now plugging in $\displaystyle u$ and $\displaystyle du$ we get:

\displaystyle \begin{align*}\int \frac{e^{6x}}{e^{6x}+9}\,\, dx=& \int \frac{1}{6u}\,\,du \\=& \frac{1}{6} \ln|u| + C\\=& \frac{1}{6}\ln| e^{6x}+9| + C.....\text{[by plugging in value of } u\text{]}\end{align*}

$\displaystyle \int \frac{e^{6x}}{e^{6x}+9}\,\, dx = \frac{1}{6}\ln| e^{6x}+9| + C.....\text{[Answer]}$

You can check the answer here:

int (e^(6x))/(e^(6x)+9) - Wolfram|Alpha

4. ## Re: U substitution question

Originally Posted by x3bnm
$\displaystyle \int \frac{e^{6x}}{e^{6x}+9}\,\, dx$

Let $\displaystyle u = e^{6x} + 9\,\,\therefore du = 6e^{6x}\,\,dx$

Now plugging in $\displaystyle u$ and $\displaystyle du$ we get:

\displaystyle \begin{align*}\int \frac{e^{6x}}{e^{6x}+9}\,\, dx=& \int \frac{1}{6u}\,\,du \\=& \frac{1}{6} \ln|u| + C\\=& \frac{1}{6}\ln| e^{6x}+9| + C.....\text{[by plugging in value of } u\text{]}\end{align*}

$\displaystyle \int \frac{e^{6x}}{e^{6x}+9}\,\, dx = \frac{1}{6}\ln| e^{6x}+9| + C.....\text{[Answer]}$

You can check the answer here:

int (e^(6x))/(e^(6x)+9) - Wolfram|Alpha
And of course, since \displaystyle \displaystyle \begin{align*} e^{6x} > 0 \end{align*} for all x, so is \displaystyle \displaystyle \begin{align*} e^{6x} + 9 \end{align*}, so the absolute value signs here are overkill