# How do you integrate int ((3+lnx)^2(2-lnx))/(4x) using substitution?

Aug 26, 2016

$18 \ln x + \frac{3}{2} {\left(\ln x\right)}^{2} - \frac{4}{3} {\left(\ln x\right)}^{3} - \frac{1}{4} {\left(\ln x\right)}^{4} + C$

#### Explanation:

We have that:

$\int \frac{{\left(3 + \ln x\right)}^{2} \left(2 - \ln x\right)}{4 x} \mathrm{dx}$

So I would say the best substitution would be:

$u = \ln x$

And from this we get :$\mathrm{du} = \frac{1}{x} \mathrm{dx}$

We can substitute this into the integral to obtain:

$\int \frac{{\left(3 + u\right)}^{2} \left(2 - u\right)}{4} \mathrm{du}$

Now expand the brackets:

$= \frac{1}{4} \int \left(9 + 6 u + {u}^{2}\right) \left(2 - u\right) \mathrm{du}$

$= \int \left(18 + 3 u - 4 {u}^{2} - {u}^{3}\right) \mathrm{du}$

Which integrates to give us:

$= 18 u + \frac{3}{2} {u}^{2} - \frac{4}{3} {u}^{3} - \frac{1}{4} {u}^{4} + C$

Finally we can reverse the substitution:

$= 18 \ln x + \frac{3}{2} {\left(\ln x\right)}^{2} - \frac{4}{3} {\left(\ln x\right)}^{3} - \frac{1}{4} {\left(\ln x\right)}^{4} + C$