Question

Tfraiz!!!!!!!!!!!!!!!!!

Answer 1

Your integral is:
\[\pi \int\limits (9+\frac{\ln^2(x)}{x^2}-6\frac{\ln(x)}{x})dx\]
Obviously the 9 and the -6ln(x)/x is easy to integrate. The middle one is the feather.
Now this is where you have to be a clever bastard (like myself) and notice something.
\[\int\limits \frac{\ln^2(x)}{x^2}dx\]
Let u=ln(x)
that means that e^u=x
Then e^udu=dx
REPLACE EVERYTHING
\[\int\limits \frac{(u^2)(e^u)du}{(e^u)^2}=\int\limits \frac{u^2 du}{e^u}=\int\limits e^{-u}u^2du\]
Now do tabular method with u=u^2 and dv=e^-u
Giving:
\[\left[\begin{matrix}u & dv \\ u^2& e^{-u}\\2u&-e^{-u}\\2&e^{-u}\\0&-e^{-u}\end{matrix}\right]\]
Then you get:
\[-u^2e^{-u}-2ue^{-u}-2e^{-u}\]
Replacing everything you have:
\[-(\ln^2(x)\frac{1}{x}+2\ln(x)\frac{1}{x}+2\frac{1}{x})=-\frac{\ln^2(x)+2\ln(x)+2}{x}\]
Which, if you plug it into wolfram, you get the same thing (though I used an "easier" method.

Answer 2

How's your brain feel?