I need help with this integral,
$$\int \ln(x^2 e^x)\, dx $$
What I have:
I know ,that I can not integer $\ln$ so I rewrite it like this:
$$\int 1 \ln(x^2 e^x)\, dx = \left| \begin{array}{cc} u=\ln(x^2 e^x) & v'=\frac{1}{x^2 e^x} \\ u'=1 & v=x \end{array} \right| = x\ln(x^2 e^x)- \int \frac{1}{x^2 e^x}x = ?$$
Now I am not sure what to do.
The result is:
$$2 (x\ln x -x)+\frac{1}{2}x^2 + C $$
Thank you very much.
$\int \ln{(x^2e^x)}$ $dx$
$=x \ln {(x^2e^x)}-\int x[\frac{2xe^x+x^2e^x}{x^2e^x}]$ $dx$
$=x\ln {(x^2e^x)}- \int (2+x)$ $dx$
$=x\ln {(x^2e^x)}-2x-\frac{x^2}{2}$
$=x\ln{x^2}+x \ln{e^x}- 2x- \frac{x^2}{2}$
$=2x\ln{x}+x^2-2x-\frac{x^2}{2}+C$
$=2(x\ln{x}-x)+ \frac{x^2}{2} +C$
You've made a mistake with your integration by parts.
Let $u=\ln(x^2 e^x)$ and $v'=1$.
Hence, $u'=\frac{x+2}{x}$ and $v=x$.
Substituing into $\int uv'= uv-\int vu'$ gives:
$$x\ln(x^2 e^x)-\int (x+2)~dx$$
Can you take it from here?
Keep in mind that $\ln(x^2 e^x)$ can be simplified by using the identity $\ln(ab)=\ln(a)+\ln(b)$ to give $\ln(x^2 e^x)=\ln(x^2)+\ln(e^x)=2\ln{|x|}+x$ to give the form of the answer similar to your book.
To do by parts, your choice of $f $ and $g $ is perfect, but for the implementation. We have $$I = x [2\ln x +x]-\int (x)(\frac {2}{x} +1) dx $$ giving us the answer as in the book.