# What is int (1+lnx)/x^2dx?

Mar 15, 2018

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

#### Explanation:

We seek:

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

In preparation for an application of [integration by Parts]We can then apply Integration By Parts:

Let  { (u,=1+lnx, => (du)/dx,=1/x), ((dv)/dx,=1/x^2, => v,=-1/x ) :}

Then plugging into the IBP formula:

$\int \setminus \left(u\right) \left(\frac{\mathrm{dv}}{\mathrm{dx}}\right) \setminus \mathrm{dx} = \left(u\right) \left(v\right) - \int \setminus \left(v\right) \left(\frac{\mathrm{du}}{\mathrm{dx}}\right) \setminus \mathrm{dx}$

We have:

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

Giving is:

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

$\setminus \setminus = - \frac{1 + \ln x}{x} - \frac{1}{x} + C$

$\setminus \setminus = - \frac{2 + \ln x}{x} C$