well, one way:
let z = 3 + 2x
then
dz = 2 dx so dx = (1/2)dz
if x = 0, z = 3
if x = 2, z = 7
so
7
∫ (5/z)(1/2) dz
3
evaluate the definite integr
2
∫ 5/(3+2x) dx
0
2 answers
2
∫ 5/(3+2x) dx
0
= [ (5/2) ln(3 + 2x) ] from 0 to 2
= (5/2)( ln 7- ln 3)
= (5/2) ln (7/3)
∫ 5/(3+2x) dx
0
= [ (5/2) ln(3 + 2x) ] from 0 to 2
= (5/2)( ln 7- ln 3)
= (5/2) ln (7/3)