Question

How do you integrate `ln(3x+4)`?

Answer 1

`I = int \ ln(3x+4) \ dx = ((3x+4)(ln(3x+4)-1))/3 + C`

Explanation:

We seek:

`I = int \ ln(3x+4) \ dx`

First we perform a substitution:

Let `t=3x+4 => (dt)/dx = 3`

Substituting into the integral we get:

`I = 1/3 \ int \ ln(3x+4) \ 3 \ dx`
`\ \ = 1/3 \ int \ ln(t) \ dt`

We can use Integration By Parts (IBP):

`int \ u(dv)/dx \ dx = uv - int \ v(du)/dx \ dx`, or less formally
`" " int \ u \ dv=uv-int \ v \ du`

Let `{ (u,=lnt, => (dt)/dx,=1/t), ((dv)/dt,=1, => v,=t ) :}`

Then plugging into the IBP formula:

`int \ ln(t)(1) \ dt = (ln(t))(t) - int \ (t)(1/t) \ dt`
`:. int \ lnt \ dt = tlnt - int \ dt`
`:. int \ lnt \ dt = tlnt - t`

Hence we have:

`I = 1/3 \ (tlnt-t) + C`

Restoring the substitution we get:

`I = 1/3 \ ((3x+4)ln(3x+4)-(3x+4)) + C`
`\ \ = ((3x+4)(ln(3x+4)-1))/3 + C`