Antiderivative of x^2*ln x^2

[tborchardt]

Hello,

I'm trying to figure out what the anti-derivative of x^2*lnx^2
I try doing integral by parts but the anti-derivative of ln x^2 is messy. And then I try doing integral by substitution but that doesn't work easy for me either.

I'm problably just missing something but any help is greatly appreciated to get me started.

Thanks in advance, it is greatly appreciated.

 

[MarkFL]

I would probably use a logarithmic property to write:

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

Now, integration by parts will be simpler.

 

[tborchardt]

I would probably use a logarithmic property to write:

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

Now, integration by parts will be simpler.

Wow, thanks. Can't believe I forgot that simple rule about the natural log. After that I solved it within a minute haha. Thanks again!

 

[Deleted member 4993]

Hello,

I'm trying to figure out what the anti-derivative of x^2*lnx^2
I try doing integral by parts but the anti-derivative of ln x^2 is messy. And then I try doing integral by substitution but that doesn't work easy for me either.

I'm problably just missing something but any help is greatly appreciated to get me started.

Thanks in advance, it is greatly appreciated.

Another way:

substitute

u = x²

du = 2x dx

dx = du/(2 * √u)

Now your problem becomes:

\(\displaystyle \frac{1}{2}\int \sqrt{u}*ln(u) du\)

complete integration by parts and substitute back.