# How do you integrate int lnx/x^5 by integration by parts method?

Jul 20, 2016

$= - \frac{1}{4 {x}^{4}} \left(\ln x + \frac{1}{4}\right) + C$

#### Explanation:

tactically, we're going to set the IBP up so that we get to bust up the $\ln x$ term by differentiating it. integrating $\ln x$ is a bad way to go.

So we say that

$\int \ln \frac{x}{x} ^ 5 \setminus \mathrm{dx} = \int \setminus \ln x \setminus \left(- \frac{1}{4}\right) \frac{d}{\mathrm{dx}} \left(\frac{1}{x} ^ 4\right) \setminus \mathrm{dx}$

$= \int \setminus \ln x \frac{d}{\mathrm{dx}} \left(- \frac{1}{4 {x}^{4}}\right) \setminus \mathrm{dx}$

now we apply IBP: $\int u v ' = u v - \int u ' v$

so we can crush that $\ln x$ term

$\int \setminus \ln x \frac{d}{\mathrm{dx}} \left(- \frac{1}{4 {x}^{4}}\right) \setminus \mathrm{dx}$

$= - \frac{1}{4 {x}^{4}} \ln x - \int \setminus \frac{d}{\mathrm{dx}} \left(\ln x\right) \cdot \left(- \frac{1}{4 {x}^{4}}\right) \setminus \mathrm{dx}$

$= - \frac{1}{4 {x}^{4}} \ln x + \frac{1}{4} \int \setminus \frac{1}{x} \cdot \left(\frac{1}{{x}^{4}}\right) \setminus \mathrm{dx}$

$= - \frac{1}{4 {x}^{4}} \ln x + \frac{1}{4} \int \setminus \frac{1}{{x}^{5}} \setminus \mathrm{dx}$

$= - \frac{1}{4 {x}^{4}} \ln x + \frac{1}{4} \left(- \frac{1}{4 {x}^{4}} + C\right)$

$= - \frac{1}{4 {x}^{4}} \left(\ln x + \frac{1}{4}\right) + C$ , et voila!

We can have a quick go the other way by saying that
$\int \ln \frac{x}{x} ^ 5 \setminus \mathrm{dx} = \int \setminus \frac{d}{\mathrm{dx}} \left(\frac{1}{2} {\ln}^{2} x\right) \left(\frac{1}{x} ^ 4\right) \setminus \mathrm{dx}$

So IBP tells us that equals

$= \frac{1}{2 {x}^{4}} {\ln}^{2} x - \int \setminus \frac{1}{2} {\ln}^{2} x \frac{d}{\mathrm{dx}} \left(\frac{1}{x} ^ 4\right) \setminus \mathrm{dx}$

$= \frac{1}{2 {x}^{4}} {\ln}^{2} x + 2 \int \setminus \frac{{\ln}^{2} x}{x} ^ 5 \setminus \mathrm{dx}$

dunno where that's going! but i'm stopping there...