I was studying integration when I came across mannipulating natural logarithms. It is known that:
$$ \int \frac{1}{x}dx = \ln{|x|} +c \forall x>0 \wedge \int \frac{1}{ax+b}dx =\frac{1}{a} {\ln{|ax+b|}}+c \forall ax+b >0$$
Reading from Cambridge International AS & A Level Mathematics 2 & 3:
Rafiu and Fausat are asked to find $\int \frac{1}{2(3x+1)} dx$
Rafiu writes: $\int \frac{1}{2(3x+1)} dx = \int \frac{1}{6x+2}dx$ $$=\frac{1}{6} \ln{(6x+2)}+c$$ Fausat writes: $\int \frac{1}{2(3x+1)} dx = \frac{1}{2} \int \frac{1}{3x+1} dx$ $$= \bigg( \frac{1}{2} \bigg) \bigg( \frac{1}{3} \bigg)\ln{(3x+1)}+c$$ $$=\frac{1}{6}\ln{(3x+1)}+c$$ Decide who is correct and dsiscuss the reasons your decision with your classmates.
I think Fausat is right... because he applied two applicable rules. Why is Rafiu wrong though? Should I always keep the denominator simplified, why so?
Both are correct, because indefinite integrals are always computed up to a constant. Indeed, in the present case, one has $\ln(6x+2) = \ln(2(3x+1)) = \ln(3x+1) + \ln(2)$, so that the two constants of integration are related by $c_{Rafiu} = c_{Fausat} + \frac{1}{6}\ln(2)$.