Find ∫ (ln x) ^2 dx.
Thank you!!!
Use integration by parts:
$\displaystyle \begin{array}{ll}u = \big[\ln(x)\big]^{2} & dv = dx \\ du = 2\ln(x) \cdot \frac{1}{x} & v = x\end{array}$
So,
$\displaystyle \int \big[\ln(x)\big]^{2} \: dx = x\big[\ln(x)\big]^{2} - \int \left(2\ln(x) \cdot \frac{1}{x} \cdot x\right) dx$
$\displaystyle \int \big[\ln(x)\big]^{2} \: dx = x\big[\ln(x)\big]^{2} - 2{\color{blue}\int \left( \ln(x)\right) dx}$
The thing in blue needs integration by parts again but I'll leave it up to you with that one (use a similar sub as we did earlier!).
This was my first thought...and I'm gonna go along with it...
$\displaystyle \int (\ln(x))^2\,dx$
let $\displaystyle u=\ln(x)\implies x=e^u$
thus, $\displaystyle du=\frac{\,dx}{x}\implies \,dx=x\,du$
Thus, $\displaystyle \int(\ln(x))^2\,dx=\int u^2e^{u}\,du$
Apply integration by parts...
$\displaystyle w=u^2$ and $\displaystyle \,dv=e^{u}\,du$
$\displaystyle \therefore \,dw=2u\,du$ and $\displaystyle v=e^{u}$
Thus, we get $\displaystyle -u^2e^{u}-2\int ue^{u}\,du$
let $\displaystyle w=u$ and $\displaystyle \,dv=e^{u}\,du$
$\displaystyle \therefore \,dw=\,du$ and $\displaystyle v=e^{u}$
Now we see that we get
$\displaystyle u^2e^{u}-2\left[ue^{u}-\int e^{u}\,du\right]=e^{u}\left(u^2-2u+2\right)$
Substitute $\displaystyle u=\ln(x)$ back in to get
$\displaystyle \int (\ln(x))^2\,dx=\color{red}\boxed{x\left[(\ln(x))^2-2\ln(x)+2\right]}$
Does this make sense?
--Chris
EDIT: o_O beat me to it!