# Math Help - Having trouble with integrals with LN, exam tomarrow

1. ## Having trouble with integrals with LN, exam tomarrow

Hi guys, i have an exam tomarrow in calc II and i have been having the same problem with integrals of ln in quotients

for example, i know through the answer key and my TI89 that

$\int
\frac{\ln(n)}{(n^2)}=\frac{-\ln(x)-1}{x}
$

and

$\int{\frac{\ln{(x+1)}}{x+1}}=\frac{(ln(x+1))^2}{2}$

and

$\int{\frac{1}{\sqrt{x}*(\sqrt{x}+1)}}=\ln{\sqrt{n} }+1$

but im really not sure how they go about getting this, i shouldnt have to use IBP or some real complicated method because its past that section (im actually in the integral test of infinite series) and even in the step wise explanation it just goes from the integral to the solution like its a simple mental integral.

Can anybody explain the shortcut they are using?

Thanks so much

2. Originally Posted by Alex01tib
Hi guys, i have an exam tomarrow in calc II and i have been having the same problem with integrals of ln in quotients

for example, i know through the answer key and my TI89 that

$\int
\frac{\ln(n)}{(n^2)}=\frac{-\ln(x)-1}{x}
$

integration by parts ... $\textcolor{red}{u = \ln{n} \, , \, dv = \frac{1}{n^2}}$

and

$\int{\frac{\ln{(x+1)}}{x+1}}=\frac{(ln(x+1))^2}{2}$

straight substitution ... $\textcolor{red}{u = \ln(x+1) \, , \, du = \frac{1}{x+1}}$

and

$\int{\frac{1}{\sqrt{x}*(\sqrt{x}+1)}}=\ln{\sqrt{n} }+1$

straight substitution again ... $\textcolor{red}{u = \sqrt{x} + 1 \, , \, du = \frac{1}{2\sqrt{x}}}$
.

3. Hey thanks so much, sometimes you just need a bit of a refresher to get back into the integration with ln game lol. I remember now haha...

Appreciate it man.