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using integration by parts, find $\displaystyle \int [ln(x)]^2 dx$

 Jun 19, 2022
 #1
avatar+117758 
+1

Hint:

the integral is lnx =   the integral of (1*lnx)  

do that first.

You need to use integration by parts twice.

 

 

I have not done it, I don't have time at present, but that is how I would proceed.

 Jun 19, 2022
 #3
avatar+117758 
+1

\displaystyle \int [ln(x)]^2 dx

 

\(\displaystyle \int [ln(x)]^2 dx\\ =\displaystyle \int [ln(x)] [ln(x)] dx\\ u=lnx\qquad v'=lnx\\ u'=\frac{1}{x}\qquad v=?\\~\\ ---------------------\\ \displaystyle \int [ln(x)] dx\\ =\displaystyle \int 1*[ln(x)] dx\\ p=lnx\qquad q'=1\\ p'=\frac{1}{x}\qquad q=x\\ so\\ \displaystyle \int 1*[ln(x)] dx\\ =xlnx-\int \frac{1}{x}*x\;dx\\ =xlnx-\int 1\;dx\\ =xlnx-x\\~\\ \text {So v=xlnx-x}\\~\\ ----------------\\ \)

Going back to the beginning

 

\(\displaystyle \int [ln(x)]^2 dx\\ =\displaystyle \int [ln(x)]  [ln(x)] dx\\ \qquad u=lnx\qquad v'=lnx\\ \qquad u'=\frac{1}{x}\qquad v=xlnx-x\\~\\ =uv-\int vu' dx \;\;+c \)

 

And you can finish it 

 

 

LaTex:

\displaystyle \int [ln(x)]^2 dx\\
=\displaystyle \int [ln(x)]  [ln(x)] dx\\
u=lnx\qquad v'=lnx\\
u'=\frac{1}{x}\qquad v=?\\~\\
---------------------\\
\displaystyle \int [ln(x)] dx\\
=\displaystyle \int 1*[ln(x)] dx\\
p=lnx\qquad q'=1\\
p'=\frac{1}{x}\qquad q=x\\
so\\
\displaystyle \int 1*[ln(x)] dx\\
=xlnx-\int \frac{1}{x}*x\;dx\\
=xlnx-\int 1\;dx\\
=xlnx-x\\~\\
\text {So v=xlnx-x}\\~\\
----------------\\

Melody  Jun 20, 2022
 #2
avatar+9458 
+1

\(\newcommand{\dint}{\displaystyle\int} \begin{array}{rcl} \dint (\ln x)^2 \,dx &=& x(\ln x)^2 - \dint x\,d\left((\ln x)^2\right)\\ &=& x(\ln x)^2 - \dint x\cdot \dfrac{2\ln x}{x}\,dx\\ &=& x(\ln x)^2 - 2\dint \ln x\,dx\\ &=& x(\ln x)^2 - 2\left(x\ln x - \dint x\,d(\ln x)\right)\\ &=& \cdots \end{array}\)

 

I will leave the rest to you.

 Jun 20, 2022

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