Question

How do you evaluate the integral `intx^nlnxdx`?

Answer 1

For `n=-1`: `(lnx)^2/2+C`

For `n!=-1`: `(x^(n+1)((n+1)lnx-1))/(n+1)^2+C`

Explanation:

`I=intx^nlnxdx`

Use integration by parts. Let:

`{(u=lnx,=>,du=1/xdx),(dv=x^ndx,=>,v=x^(n+1)/(n+1)):}" "" "" "(n!=-1)`

Then:

`I=lnx(x^(n+1)/(n+1))-intx^(n+1)/(n+1)(1/x)dx`

`color(white)I=x^(n+1)/(n+1)lnx-1/(n+1)intx^ndx`

`color(white)I=x^(n+1)/(n+1)lnx-1/(n+1)(x^(n+1)/(n+1))`

`color(white)I=(x^(n+1)((n+1)lnx-1))/(n+1)^2+C`

In the case where `n=-1`, the integral is `intlnx/xdx`. With the substitution `t=lnx` this becomes `inttdt=t^2/2=(lnx)^2/2+C`.