Question #5f659

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Answer 1 · 2017-03-14T00:44:10

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

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#