Calculus: Integration problem?

Question

What is the indefinite integral of:

int: u^3/(2u-1)^3 du

Thanks for any tips, help!

Answer 1

let v = 2u -1, so u = (1/2)(v+1)

then dv = 2du, or du = (1/2)dv

then your integral becomes:

∫u^3/(2u-1)^3 du = (1/16)∫ (v+1)^3/v^3 dv

=(1/16) ∫ (v^3 + 3v^2 + 3v + 1)/v^3 dv

=(1/16) ∫1+(3/v) + (3/v^2) + (1/v^3) dv

from here it is just normal polynomal integration.  When you are done, substitute 2u-1 back in for all the v's

Answer 2

∫ [u³ / (2u - 1)³] du =

let (2u - 1) = t →

2u = t + 1 →

u = (1/2)(t + 1)

du = (1/2) dt

then substitute, yielding:

∫ [u³ / (2u - 1)³] du = ∫ {[(1/2)(t + 1)]³ / t³ } (1/2) dt =

∫ {[(1/8)(t + 1)³] / t³ } (1/2) dt =

(1/8)(1/2) ∫ [(t + 1)³ / t³] dt  =

expand the numerator:

(1/16) ∫ [(t³ + 3t² + 3t + 1) / t³] dt =

and distribute it as:

(1/16) ∫ [(t³/ t³) + (3t²/ t³) + (3t/ t³) + (1/ t³)] dt =

(1/16) ∫ (t³/ t³) dt + (1/16) ∫ (3t²/ t³) dt + (1/16) ∫ (3t/ t³) dt + (1/16) ∫ (1/ t³) dt =

simplify an take the constants out:

(1/16) ∫ dt + (3/16) ∫ (1/ t) dt + (3/16) ∫ (1/ t²) dt + (1/16) ∫ (1/ t³) dt =

express the integrands as negative powers:

(1/16) ∫ dt + (3/16) ∫ (1/ t) dt + (3/16) ∫ t^(-2) dt + (1/16) ∫ t^(-3) dt =

(1/16) t + (3/16) ln | t | + (3/16) [t^(-2+1)]/(-2+1) + (1/16) [t^(-3+1)] /(-3+1) + C =

(1/16) t + (3/16) ln | t | + (3/16) [t^(-1)]/(-1) + (1/16) [t^(-2)] /(-2) + C =

(1/16) t + (3/16) ln | t | - (3/16) (1/ t) - (1/32) (1/ t²) + C

then substitute back t = (2u - 1), yielding:        

∫ [u³ / (2u - 1)³] du =

(1/16)(2u -1) + (3/16) ln |2u - 1| - (3/16) [1/(2u -1)] - (1/32) [1/ (2u -1)²] + C

I hope it helps...

Bye!