# How do you integrate f(x)=(1+lnx)^4/x using the quotient rule?

Sep 11, 2017

$\int \frac{{\left(1 + \ln x\right)}^{4}}{x} \mathrm{dx} = \frac{1}{5} {\left(1 + \ln x\right)}^{5} + C$

#### Explanation:

There is no quotient rule for integration. Instead we will have to approach this another way.

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

let's see what happens when we differentiate ${\left(1 + \ln x\right)}^{5}$

$y = {\left(1 + \ln x\right)}^{5}$

by the chain rule

$u = 1 + \ln x \implies \frac{\mathrm{du}}{\mathrm{dx}} = \frac{1}{x}$

$y = {u}^{5} \implies \frac{\mathrm{dy}}{\mathrm{du}} = 5 {u}^{4} = 5 {\left(1 + \ln x\right)}^{4}$

$\therefore \frac{\mathrm{dy}}{\mathrm{dx}} = \frac{\mathrm{dy}}{\mathrm{du}} \frac{\mathrm{du}}{\mathrm{dx}}$

$\frac{\mathrm{dy}}{\mathrm{dx}} = 5 {\left(1 + \ln x\right)}^{4} \times \frac{1}{x} = \frac{5 {\left(1 + \ln x\right)}^{4}}{x}$

we notice that comparing this with the integral it is same except for the constant, so we conclude:

$\int \frac{{\left(1 + \ln x\right)}^{4}}{x} \mathrm{dx} = \frac{1}{5} {\left(1 + \ln x\right)}^{5} + C$