Find the antiderivative: Polynomial over polynomial

Question

$$\int \frac{x-1}{x^2 + 3x}\;dx $$

After working through it, I arrive at

$$ \frac{1}{3} \Big(\,4\log \big( \,x+3\,\big) - \log \left(\,x \,\right) \Big)$$

Which webwork deems incorrect. Am I going wrong, here?

Answer 1

Notice, $$\int \frac{x-1}{x^2+3x}$$ $$=\int \frac{x-1}{x(x+3)}dx$$ Using partial fractions $$=\int\left( -\frac{1}{3x}+\frac{4}{3(x+3)}\right)dx$$ $$=-\frac{1}{3}\int \frac{dx}{x}+\frac{4}{3}\int \frac{dx}{x+3}$$ $$=-\frac{1}{3}\ln|x|+\frac{4}{3}\ln|x+3|+C$$ $$=\color{red}{\frac{1}{3}(4\ln|x+3|-\ln|x|)+C}$$

You need to use mode '$|\ \ |$' instead of parenthesis '$(\ \ )$' & add a constant $(C)$ of integration in your answer

Answer 2

Undisputably,

$$ \frac{1}{3} \Big(\,4\log \big( \,x+3\,\big) - \log \left(\,x \,\right) \Big)'=\frac{1}{3} \Big(\frac4{x+3} - \frac1x \Big)=\frac{x-1}{(x+3)x}.$$ WeBWorK seems wrong to tell you wrong (requirements to use of absolute values and add a constant are debatable).