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Integration by parts

Yankel

Active member
Jan 27, 2012
398
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
Jan 26, 2012
995
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

Active member
May 31, 2013
118
\(\displaystyle \int\frac{3x}{3x+1}dx=\int1-\frac{1}{3x+1}dx\)
\(\displaystyle \int\frac{1}{x}dx=\text{log}x+c\)
Hope that helps.:)
p.s:Sorry,I didn't see chrisL replying.(Emo)
 

chisigma

Well-known member
Feb 13, 2012
1,704
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
Jan 26, 2012
1,403
\(\displaystyle
\(\displaystyle \int\frac{1}{x}dx=\text{log}x+c\)\)
\(\displaystyle

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