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)**