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.
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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.