Integration by Parts: Solving \(\int ln(3x+1)dx\) and \(\int (ln(x))^{2}dx\)

[Yankel]

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]

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]

$$\int\frac{3x}{3x+1}dx=\int1-\frac{1}{3x+1}dx$$
$$\int\frac{1}{x}dx=\text{log}x+c$$
Hope that helps.:)
p.s:Sorry,I didn't see chrisL replying.(Emo)

 

[chisigma]

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/#post24093Kind regards

$\chi$ $\sigma$

 

[Prove It]

$$
$$\int\frac{1}{x}dx=\text{log}x+c$$

No, \displaystyle \begin{align*} {\frac{1}{x}\,dx} = \log{|x|} + C \end{align*}.