QUESTION: Integrate by parts the function 9x(ln(x))^2 wrt x between x = 1 and x = e^4.
I = int 9x(ln(x))^2 dx
let u = ln(x). When x = 1, u = 0, when x = e^4, u = 4
Then x = e^u
and dx = e^u.du
So now,
I = int 9e^u.u^2.e^u du
I = 9*int e^(2u).u^2 du
integrating by parts,
I/9 = (1/2)e^(2u).u^2 - int e^(2u).u du
I/9 = (1/2)e^(2u).u^2 - {(1/2)e^(2u).u - int (1/2)e^(2u) du}
I/9 = (1/2)e^(2u).u^2 - {(1/2)e^(2u).u - (1/4)e^(2u)}
I/9 = (1/4)e^(2u){2u^2 - 2u + 1}
I = (9/4)e^(2u){2u^2 - 2u + 1}
Putting in the limits, u = 0 to u = 4
I = (9/4){[e^(0){0 - 0 + 1}] - [e^(8){32 - 8 + 1}]}
I = (9/4){[1] - [e^8*25]}
I = (9/4)(25.e^8 - 1)