Question

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

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

Answer 1

`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`

Answer 2

`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`