Question

Can you evaluate `int \ (xln(x+1))/(x+1) \ dx`?

Answer 1

I tried this:

Explanation:

Have a look:

Answer 2

`int \ (xln(x+1))/(x+1) \ dx = (x+1)ln(x+1) - x - 1/2 ln^2(x+1) + C`

Explanation:

Yes I can!

We want to find:

`I = int \ (xln(x+1))/(x+1) \ dx`

An obvious logical substitution would be:

Let `w=x+1 = > (dw)/dx = 1`, and `x=w=1`

Substituting into the integral gives us:

`I = int \ ((w-1)lnw)/w \ dw`
`\ \ = int \ (w lnw - ln w)/w \ dw`
`\ \ = int \ lnw - ln w/w \ dw`
`\ \ = int \ lnw \ dw - int \ ln w/w \ dw`

The first integral is a well known result and one that probably should be memorised. It can be derived using Integration by parts:

Let `{ (u,=lnw, => , (du)/(dw)=1/w), ((dv)/(dw),=1, =>, v=w ) :}`

Then plugging into the IBP formula:

`int \ (u)((dv)/dx) \ dx = (u)(v) - int \ (v)((du)/dx) \ dx`

gives us:

`int \ (lnw)(1) \ dw = (lnw)(w) - int \ (w)(1/1) \ dw`
`:. int \ lnw \ dw = wlnw - int \ dw`
`:. " " = wlnw - w`

And for the second integral we can perform another substitution:

Let `z=lnw => (dz)/(dw) = 1/w`

Substituting into the second integral gives us:

`int \ ln w/w \ dw = int \ z \ dz`
`" " = 1/2z^2`

And restoring the substitution for `w` we get

`int \ ln w/w \ dw = 1/2 (lnw)^2`

Combining these two results then gives us:

`I = {wlnw - w} - {1/2 (lnw)^2} + C_1`

And restoring the substitution for `w` gives us:

`I = (x+1)ln(x+1) - (x+1) - 1/2 ln^2(x+1) + C_1`
`\ \ = (x+1)ln(x+1) - x-1 - 1/2 ln^2(x+1) + C_1`
`\ \ = (x+1)ln(x+1) - x - 1/2 ln^2(x+1) + C`