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:

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

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

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