Question

Question #5f659

Answer 1

`2sqrt(1+lnx)+lnabs((sqrt(1+lnx)-1)/(sqrt(1+lnx)+1))+C`

Explanation:

Use the substitution `u=1+lnx`. This implies that `du=1/xdx` and `lnx=u-1`. Then:

`intsqrt(1+lnx)/(xlnx)dx=intsqrtu/(u-1)du`

Now let `v=sqrtu`. This implies that `v^2=u`, so `(2v)dv=du`.

`=intv/(v^2-1)(2v)dv=2intv^2/(v^2-1)dv`

Rewriting the integrand as `(v^2-1+1)/(v^2-1)=1+1/(v^2-1)`:

`=2int(1+1/(v^2-1))dv`

We can perform partial fraction decomposition on `1/(v^2-1)=1/((v+1)(v-1))`:

`1/(v^2-1)=A/(v+1)+B/(v-1)`

`1=A(v-1)+B(v+1)`

Letting `v=1` reveals that `B=1/2` and letting `v=-1` reveals that `A=-1/2`. Thus

`1/(v^2-1)=1/(2(v-1))-1/(2(v+1))`

Then the integral becomes:

`=2int(1+1/(2(v-1))-1/(2(v+1)))dv`

`=2intdv+int(dv)/(v-1)-int(dv)/(v+1)`

These are simple integrals to find:

`=2v+ln(abs(v-1))-ln(abs(v+1))+C`

`=2sqrtu+lnabs((sqrtu-1)/(sqrtu+1))+C`

`=2sqrt(1+lnx)+lnabs((sqrt(1+lnx)-1)/(sqrt(1+lnx)+1))+C`