Integration

**Tangera** · August 10th 2008, 09:48 PM

Find ∫ (ln x) ^2 dx.

Thank you!!!

**o_O** · August 10th 2008, 10:10 PM

Use integration by parts:

$\begin{array}{ll}u = \big[\ln(x)\big]^{2} & dv = dx \\ du = 2\ln(x) \cdot \frac{1}{x} & v = x\end{array}$

So,
$\int \big[\ln(x)\big]^{2} \: dx = x\big[\ln(x)\big]^{2} - \int \left(2\ln(x) \cdot \frac{1}{x} \cdot x\right) dx$
$\int \big[\ln(x)\big]^{2} \: dx = x\big[\ln(x)\big]^{2} - 2{\color{blue}\int \left( \ln(x)\right) dx}$

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!).

**Chris L T521** · August 10th 2008, 10:16 PM

Originally Posted by Tangera

Find ∫ (ln x) ^2 dx.

Thank you!!!

This was my first thought...and I'm gonna go along with it...

$\int (\ln(x))^2\,dx$

let $u=\ln(x)\implies x=e^u$

thus, $du=\frac{\,dx}{x}\implies \,dx=x\,du$

Thus, $\int(\ln(x))^2\,dx=\int u^2e^{u}\,du$

Apply integration by parts...

$w=u^2$ and $\,dv=e^{u}\,du$

$\therefore \,dw=2u\,du$ and $v=e^{u}$

Thus, we get $-u^2e^{u}-2\int ue^{u}\,du$

let $w=u$ and $\,dv=e^{u}\,du$

$\therefore \,dw=\,du$ and $v=e^{u}$

Now we see that we get

$u^2e^{u}-2\left[ue^{u}-\int e^{u}\,du\right]=e^{u}\left(u^2-2u+2\right)$

Substitute $u=\ln(x)$ back in to get

$\int (\ln(x))^2\,dx=\color{red}\boxed{x\left[(\ln(x))^2-2\ln(x)+2\right]}$

Does this make sense?

--Chris

EDIT: o_O beat me to it!

**Chris L T521** · August 10th 2008, 10:18 PM

Originally Posted by o_O

Use integration by parts:

$\begin{array}{ll}u = \big[\ln(x)\big]^{2} & dv = dx \\ du = 2\ln(x) \cdot \frac{1}{x} & v = x\end{array}$

So,
$\int \big[\ln(x)\big]^{2} \: dx = x\big[\ln(x)\big]^{2} - \int \left(2\ln(x) \cdot \frac{1}{x} \cdot x\right) dx$
$\int \big[\ln(x)\big]^{2} \: dx = x\big[\ln(x)\big]^{2} - 2{\color{blue}\int \left( \ln(x)\right) dx}$

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!).

Why do I have to make life so hard?

--Chris

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