Question #bc6c7

Question

2569 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-08-27T10:35:58

Integrate by parts and exploit the fact the integral repeats itself. Answer is:

$I=1/4ln^2x+c$

The integral is equal to:

$I=intlnsqrt(x)/xdx$

Use $sqrt(x)=x^(1/2)$ and $lnx^a=alnx$

$I=intlnx^(1/2)/xdx=int1/2lnx/xdx=1/2intlnx*1/xdx=$

$=1/2intlnx*(lnx)'dx$

$I=1/2((lnx*lnx)-int(lnx)'*lnxdx)=$

$=1/2(ln^2x-int1/xlnxdx)=1/2ln^2x-1/2intlnx/xdx=$

$=1/2ln^2x-intlnx^(1/2)/xdx=1/2ln^2x-intlnsqrt(x)/xdx$

Noticing the last integral is equal to the original:

$I=intlnsqrt(x)/xdx$

Therefore:

$I=1/2ln^2x-I$

$I+I=1/2ln^2x$

$2I=1/2ln^2x$

$I=1/4ln^2x$

Adding the constant of integration:

$I=1/4ln^2x+c$