# How do you integrate int(lnx)^2/x by integration by parts method?

Feb 3, 2017

$\int {\left(\ln x\right)}^{2} / x \mathrm{dx} = {\left(\ln x\right)}^{3} / 3 + C$

#### Explanation:

Integration by parts is not the best way to solve this integral.
As $d \left(\ln x\right) = \frac{\mathrm{dx}}{x}$ we can substitute:

$y = \ln x$

$\mathrm{dy} = \frac{\mathrm{dx}}{x}$

and have:

$\int {\left(\ln x\right)}^{2} / x \mathrm{dx} = \int {y}^{2} \mathrm{dy} = {y}^{3} / 3 + C = {\left(\ln x\right)}^{3} / 3 + C$

Feb 3, 2017

I got: $\frac{1}{3} {\left(\ln \left(x\right)\right)}^{3} + c$

#### Explanation:

We have:
$\int {\left(\ln \left(x\right)\right)}^{2} / x \mathrm{dx} = \int \frac{1}{x} {\left(\ln \left(x\right)\right)}^{2} \mathrm{dx} =$ by parts:
$\ln \left(x\right) {\left(\ln \left(x\right)\right)}^{2} - \int \ln \left(x\right) \cdot 2 \ln \left(x\right) \cdot \frac{1}{x} \mathrm{dx} =$
$= {\left(\ln \left(x\right)\right)}^{3} - 2 \int {\left(\ln \left(x\right)\right)}^{2} / x \mathrm{dx}$
so that basically we have that:

$\int {\left(\ln \left(x\right)\right)}^{2} / x \mathrm{dx} = {\left(\ln \left(x\right)\right)}^{3} - 2 \int {\left(\ln \left(x\right)\right)}^{2} / x \mathrm{dx}$

now a trick...
take the last integral to the left of the equal sign as in a normal equation:
$\int {\left(\ln \left(x\right)\right)}^{2} / x \mathrm{dx} + 2 \int {\left(\ln \left(x\right)\right)}^{2} / x \mathrm{dx} = {\left(\ln \left(x\right)\right)}^{3}$
add the two integrals and rearrange:
$3 \int {\left(\ln \left(x\right)\right)}^{2} / x \mathrm{dx} = {\left(\ln \left(x\right)\right)}^{3}$
$\int {\left(\ln \left(x\right)\right)}^{2} / x \mathrm{dx} = \frac{1}{3} {\left(\ln \left(x\right)\right)}^{3} + c$

Feb 3, 2017

$\int {\left(\ln x\right)}^{2} / x \mathrm{dx} = \frac{1}{3} {\left(\ln x\right)}^{3} + C$

#### Explanation:

$I = \int {\left(\ln x\right)}^{2} / x \mathrm{dx}$

Integration by parts is not necessary. The quickest way to do this is with the substitution $u = \ln x$ which implies that $\mathrm{du} = \frac{1}{x} \mathrm{dx}$. Then:

$I = \int {\left(\ln x\right)}^{2} \left(\frac{1}{x} \mathrm{dx}\right) = \int {u}^{2} \mathrm{du} = \frac{1}{3} {u}^{3} = \frac{1}{3} {\left(\ln x\right)}^{3} + C$

We can do integration by parts, however, letting:

$\left\{\begin{matrix}u = {\left(\ln x\right)}^{2} & \implies & \mathrm{du} = \frac{2 \ln x}{x} \mathrm{dx} \\ \mathrm{dv} = \frac{1}{x} \mathrm{dx} & \implies & v = \ln x\end{matrix}\right.$

Then:

$I = u v - \int v \mathrm{du}$

$I = {\left(\ln x\right)}^{3} - 2 \int {\left(\ln x\right)}^{2} / x \mathrm{dx}$

This is the original integral:

$I = {\left(\ln x\right)}^{3} - 2 I$

$3 I = {\left(\ln x\right)}^{3}$

$I = \frac{1}{3} {\left(\ln x\right)}^{3} + C$