Question

Question #bc6c7

Answer 1

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

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

Explanation:

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`

Use integration by parts:

`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`