Integration problem

[mooshupork34]

I am studying for a math test, and I was wondering how one would go about integrating the function below whether by substitution or parts (either method is fine):

ln^2(x)


It reads ln squared of x in case that looks confusing.
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[galactus]

Is that \(\displaystyle \L\\(ln(x))^{2}\) or \(\displaystyle \L\\ln(x^{2})\)?.

I've never seen \(\displaystyle \L\\ln^{2}(x)\)

 

[pka]

It can be done by parts.
\(\displaystyle \begin{array}{l}
u = \ln ^2 (x)\quad \& \quad dv = dx \\
\int {\ln (x) = x\ln (x) - x} \\
\end{array}.\)


P.S. galactus have you ever seen \(\displaystyle \sin ^2 (x)\)?

 

[mooshupork34]

thanks! I'm confused, though...what happens to the squared that comes after the ln?

 

[galactus]

It can be done by parts.
\(\displaystyle \begin{array}{l}
u = \ln ^2 (x)\quad \& \quad dv = dx \\
\int {\ln (x) = x\ln (x) - x} \\
\end{array}.\)


P.S. galactus have you ever seen \(\displaystyle \sin ^2 (x)\)?

Of course. I have not seen that used with ln, though. I assume it's the same as \(\displaystyle (ln(x))^{2}\).

 

[pka]

This is just a matter of historical notation.
If \(\displaystyle f\) is a function then \(\displaystyle f^2\) denotes the square of the function.
Whereas \(\displaystyle f^{(2)}\) denotes the second derivative of the function.

 

[mooshupork34]

ok, so i did the problem like this.

u = ln squared of x
v = x
v' = 1
u = 2ln(x)/x


Then I did:

ln squared of x * x - the integral of [2ln(x)/x] *[x/1]

This led to:

x ln squared of x - the integral of 2ln(x)
From this I got x ln squared of x - 2x(ln(x)-1)

Is this right? Or did I do this completely wrong?

 

[pka]

\(\displaystyle \L\begin{array}{l}
u = \ln ^2 (x)\quad \& \quad dv = dx \\
du = \frac{{2\ln (x)}}{x}dx\quad \& \quad v = x \\
x\ln ^2 (x) - \int {2\ln (x)dx} \\
x\ln ^2 (x) - \left[ {2\left( {x\ln (x) - x} \right)} \right] \\
\end{array}.\)

 

[soroban]

Hello, mooshupork34!

pka is absolutely correct . . .

\(\displaystyle \L\int(\ln x)^2\,dx\)

Integrate by parts . . .

. . \(\displaystyle \begin{array}u\:=\\ln x)^2\;& \;dv\:=\:dx \\
du \:=\:2(\ln x)\cdot\frac{dx}{x}\;& \;v\:=\:x\end{array}\)

We have: \(\displaystyle \L\:x\cdot(\ln x)^2\:-\:2\int \ln x\,dx\)

. . By parts again.

. . . . \(\displaystyle \begin{array}{cc}u\:=\:\ln x\;& \;dv\:=\:dx \\
du \:=\:\frac{dx}{x}\;& \;v\:=\:x\end{array}\)

We have: \(\displaystyle \L\:x(\ln x)^2\:-\:2\left[x\cdot\ln x\,-\,\int dx\right]\)

. . . . . \(\displaystyle \L= \;x(\ln x)^2\:-\:2x(\ln x)\:+\:2\int dx\)

. . . . . \(\displaystyle \L= \;x(\ln x)^2\:-\:2x(\ln x)\:+\;2x\:+\:C\)