Question

How do you evaluate the integral `int ln(3x+4)`?

Answer 1

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

Explanation:

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

N.B. Since `I` is an indefinite integral it cannot be "evaluated". Rather it can be expressed as a function of `x` plus an arbitrary constant.

Let `u=(3x+4) -> dx = (du)/3`

`:. I= int (ln u)/3 du = 1/3 int lnu du`

Integration by Parts states: `int f(x)g'(x) dx= f(x)g(x) - int f'(x)g(x) dx`

Considering `3I = int lnu * 1 du`

`f(x) = lnu , g'(x) = 1`

Thus, `f'(x) = 1/u, g(x) =u`

`:. 3I = lnu*u - int (1/u *u)du`

`= u lnu - int 1 du`

`= u lnu -u`

`:. I = u/3( lnu -1) +C`

Undo substitution.

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

Answer 2

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

Explanation:

We got: `intln(3x+4) \ dx`

Let's use u-substitution. Let `u=3x+4,:.du=3 \ dx,dx=(du)/3`

And so,

`=intlnu \ (du)/3`

Taking out the constant, the problem becomes,

`=1/3intlnu \ du`

Let's find `intlnu \ du`. We got:

`intlnu \ du`

`=intlnu*1 \ du`

Now, we use integration by parts, which is:

`intu \ dv=uv-intv \ du`

Let `u=lnu,dv=1`.

`:.v=u,du=1/u`

And so,

`intlnu*1 \ du=u\lnu-intu*1/u \ du`

`=u\lnu-int1 \ du`

`=u\lnu-u+C`

Now, we plug that back into the original integral.

We get:

`=1/3(u\lnu-u)`

Notice how I don't put in the constant yet, as we always put the constant after the final calculation, and not during the calculation, as `1/3C` would be incorrect.

`=1/3u\lnu-1/3u`

Substitute back `u=3x+4`, we get:

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

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

Now, we can add the constant.

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