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

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Discussion Overview

The discussion revolves around solving the integrals \(\int \ln(3x+1)dx\) and \(\int (\ln(x))^{2}dx\), focusing on the application of integration by parts. Participants explore various approaches and substitutions to tackle these integrals.

Discussion Character

  • Technical explanation
  • Mathematical reasoning
  • Homework-related

Main Points Raised

  • One participant proposes using integration by parts with \(u=\ln(3x+1)\) and \(v'=1\), leading to the expression \(x \cdot \ln(3x+1) - \int \frac{3x}{3x+1}dx\), and seeks guidance on the next steps.
  • Another participant suggests a substitution approach for \(\int \frac{3x}{3x+1}dx\) by rewriting it as \(\int 1 - \frac{1}{3x+1}dx\) and indicates that this could simplify the integral.
  • One participant mentions that integration by parts is necessary for \(\int (\ln(x))^{2}dx\) and recommends letting \(u=(\ln x)^{2}\) and \(dv=dx\).
  • A different participant proposes a substitution \(u = 1 + 3x\) for \(\int \ln(3x+1)dx\) to potentially simplify the integral further.
  • There is a correction regarding the integral \(\int \frac{1}{x}dx\), with one participant emphasizing the absolute value in the logarithm expression.

Areas of Agreement / Disagreement

Participants express various methods and approaches to solve the integrals, but there is no consensus on a single method or solution. Multiple competing views and techniques remain present in the discussion.

Contextual Notes

Some participants' approaches depend on specific substitutions or manipulations that may not be universally applicable. The discussion includes unresolved steps in the integration process.

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

$\chi$ $\sigma$
 
mathworker said:
$$
$$\int\frac{1}{x}dx=\text{log}x+c$$

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

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