How to evaluate ∫ [(x)/(ax+b)] dx?

Question

Please verify my answer because I'm not sure if I'm right, I got:

{[ax + b - bln(ax + b)]/[a^2]} + C

Answer 1

As the degree of the numerator equals that of the denominator, you first divide to get a fraction with no x in the numerator.

x / (ax + b) = [ (1 / a)(ax + b) - (b / a) ] / (ax + b)

= 1 / a - (b / a) / (ax + b).

The integral is:

x / a - (b / a)(1 / a) ln | ax + b | + c

= x / a - (b / a^2) ln | ax + b | + c.

Your answer is the same, except that you don't have the modulus round the logarithm. The b / a^2 constant term can be incorporated into the arbitrary C.