How do you integrate [ln(r+1)]^2 / r+1 ?

Answer 1

use substitution
u=ln(r+1)
du = 1/(r+1) dr
so
integral of [ln(r+1)]^2 / (r+1 ) dr = integral of u^2 du = u^3 /3 + C
= (ln(r+1))^3 /3 +C .

Answer 2

∫[ln(r+1)]² / r+1 dr =

u = ln(r+1)
du = 1/(r+1) dr

=>
∫[ln(r+1)]^2 / r+1 dr =∫u² dr
∫[ln(r+1)]^2 / r+1 dr = (1/3) u³ + c

∫[ln(r+1)]^2 / r+1 dr = (1/3)*[ln(r+1)]³ + c

₢

Answer 3

put ln(r+1)=t
1/r+1=dt
t^2dt on integration gives t^3/3+C
reverting back tothe original variable
=[ln(r+1)]^3/3+C