use integration by parts to show that this function:

function to integrate: 9x(ln(x))^2 dx

Variable: x

Upper Limit: e^4

Lower Limit: 1

(9/4)(25(e^8)-1)

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)

by Level 11 User (81.5k points)
thank you, is it possible to do this calculation without substituting in u and just going straight to integration by parts? @fermat
You certainly can.
I = int 9x.(ln(x))^2 dx = (9x^2/2).(ln(x))^2 - int((9x^2/2).[2.ln(x)/x] dx

the last integral above simplifies, and you can integrate by parts again, similar to the 1st line above.

Given:  ʃ9x[lnx]^2dx lower limit; 1 and upper limit; e^4

=9ʃx [lnx]^2dx

let ...... u = [lnx]^2 ..> du/dx = 2/xlnx

and dv = xdx ..> v = x^2/2

ʃ9x[lnx]^2dx = 9{x^2/2(lnx) ^2 - ʃ2/x(lnx)*x^2/2 dx}

=9{x^2/2(lnx) ^2 - ʃx(lnx)dx}

=9{x^2/2(lnx) ^2 - x^2/2(lnx) + x^2/4

introducing the upper limits                                                        introducing the lower limits

=9{e^8/2*16 - e^8/2*4 + e^8/4}                                                 9{ 1/4 }

=9{8e^8 - 2e^8 + e^8/4}

=9{25e^8/4}

therefore

ʃ9x[lnx]^2dx =9{25e^8/4 - 9/4} =(9/4){25e^8 - 1}

by Level 3 User (4.0k points)