Section 5.3
Can somebody verify this solution for me? Thanks!
Integrate $\displaystyle\int \dfrac{8}{8x-7}dx$ using $u$ substitution and the log rule
So in general, $\displaystyle\int \dfrac{1}{u}du$ = $\ln|u|+C$
This is an integration identity that you need to know. So, this is kind of one of the weird things in math... and thus $\dfrac{1}{u}=u^{-1}$, so we have $\displaystyle\int u^{-1} du = \ln|u|+C$, which is kind of weird because it doesn't obey the general power rule that we developed that $\displaystyle\int u^b du = \dfrac{u^{b+1}}{b+1}$... It's the one except to the power rule for integration!!
Anyway, the problem is to evaluate $\displaystyle\int \dfrac{8}{8x-7}dx$. Let $u=8x-7$. Then $\dfrac{du}{dx}=8$ and so $\dfrac{du}{8}=dx$.
Thus we have:
$\displaystyle\int \dfrac{8}{8x-7}dx$
$= \displaystyle\int \dfrac{8}{u}\frac{du}{8}$
$= \displaystyle\int \dfrac{1}{u} du$
$= \ln|u| + C$
$= \ln|8x-7| + C$
edit: If you want to see why $\int \frac{1}{x}dx = ln|x| +C$, click the link that an4s linked in the comments and read "Mike's" response.