(ln^2 x)/x integral from 1 to e^2 is?

int_1^(e^2)ln^2x/xdx =?

2 Answers
Jun 24, 2018

I=8/3

Explanation:

We want to evaluate

I=int_1^(e^2)ln^2(x)/xdx

Make a substitution color(blue)(u=ln(x)=>du=1/xdx

New limits color(blue)(x=1=>u=0 and color(blue)(x=e^2=>u=2

I=int_0^2 u^2/x*xdu=int_0^2 u^2du=1/3[u^3]_0^2=8/3

Jun 24, 2018

8/3

Explanation:

to integrate int_1^(e^2)ln^2x/xdx first
substitute u=lnx so (du)/dx=1/x therby dx=xdu and x=e^u
plugging this in will give:
int_0^2u^2/e^u*e^udu the e^u cancels out and we get:
int_0^2u^2du now we can use power rule to get:
u^3/3|_0^2 now we can resubstitute u=lnx
and we will get: ln^3x/3|_1^(e^2) which we can calculate as follows: ln^3(e^2)/3-ln^3(1)/3=8/3-0/3=8/3