# Integration by parts

#### Yankel

##### Active member
Hello

I am working on this integral.

$\int ln(3x+1)dx$

I choose u=ln(3x+1) and v'=1. It got me to:

$x\cdot ln(3x+1)-\int \frac{3x}{3x+1}dx$

What should I do from here ? Some substitution ? The x up there bothers me...no easier way ? Thank!

I also need to solve:

$\int (ln(x))^{2}dx$

any hints ?

P.S must use parts on this one

#### Chris L T521

##### Well-known member
Staff member
Hello

I am working on this integral.

$\int ln(3x+1)dx$

I choose u=ln(3x+1) and v'=1. It got me to:

$x\cdot ln(3x+1)-\int \frac{3x}{3x+1}dx$

What should I do from here ? Some substitution ? The x up there bothers me...no easier way ? Thank!
Note that
$\int\frac{3x}{3x+1}\,dx = \int\frac{3x+1-1}{3x+1}\,dx = \int\frac{3x+1}{3x+1} - \frac{1}{3x+1}\,dx = \int 1-\frac{1}{3x+1}\,dx$

I'd assume you can take things from here?

I also need to solve:

$\int (ln(x))^{2}dx$

any hints ?

P.S must use parts on this one
You are correct, parts is necessary for this one. In this case, see how things go when you let $u=(\ln x)^2$ and $\,dv=\,dx$.

I hope this helps!

#### mathworker

##### Well-known member
$$\displaystyle \int\frac{3x}{3x+1}dx=\int1-\frac{1}{3x+1}dx$$
$$\displaystyle \int\frac{1}{x}dx=\text{log}x+c$$
Hope that helps. p.s:Sorry,I didn't see chrisL replying. #### chisigma

##### Well-known member
Hello

I am working on this integral.

$\int ln(3x+1)dx$

I choose u=ln(3x+1) and v'=1. It got me to:

$x\cdot ln(3x+1)-\int \frac{3x}{3x+1}dx$

What should I do from here ? Some substitution ? The x up there bothers me...no easier way ? Thank!

I also need to solve:

$\int (ln(x))^{2}dx$

any hints ?

P.S must use parts on this one
May be that the substitution $u = 1 + 3x$ that leads to the indefinite integral...

$\displaystyle 3\ \int \ln u\ du\ (1)$

... is better. The (1) and the indefinite integral...

$\displaystyle \int \ln^{2} x\ dx\ (2)$

... can be found with the general formula described in...

http://www.mathhelpboards.com/f49/integrals-natural-logarithm-5286/#post24093

Kind regards

$\chi$ $\sigma$

#### Prove It

##### Well-known member
MHB Math Helper
$$\displaystyle \(\displaystyle \int\frac{1}{x}dx=\text{log}x+c$$\)
\displaystyle No, \displaystyle \begin{align*} {\frac{1}{x}\,dx} = \log{|x|} + C \end{align*}.