Find the definite integral Find the definite integral
$$\int_{1}^{2}\frac{(3x-1)(2x+3)}{x} dx$$
I have come to an answer of $16 - \ln(8)$ which I think is very wrong..
First used integration by  parts 
 A: $$\int_{1}^{2}\frac{6x^2+7x-3}{x}dx$$
$$=\int_{1}^{2}\left(6x+7-\frac{3}{x}\right)dx$$
$$=\left(3x^2+7x-3\ln |x|\right)_{1}^{2}$$
$$=\left(3(2)^2+7(2)-3\ln |2|-(3(1)^2+7(1)-3\ln |1|)\right)$$
$$=16-3\ln 2$$ $$=16-\ln 2^3=16-\ln 8$$
A: $\displaystyle \int_{1}^{2}\frac{(3x-1)\cdot (2x+3)}{x}dx = \int_{1}^{2}\frac{6x^2+7x-3}{x}dx = \int_{1}^{2}6xdx+\int_{1}^{2}7dx-\int_{1}^{2}\frac{1}{x}dx$
$\displaystyle = [3x^2]_{1}^{2}+[7x]_{1}^{2}-[3\ln|x|]_{1}^{2} = 16-3\ln(2)$
A: expanding the integrad and dividing by $x$ we get $$\int_{1}^{2}7+6x-\frac{3}{x}dx$$
the indefinite integral is given by $$7x+3x^2-3\ln|x|+C$$
A: If I'm interpreting your question correctly, your answer is right.  A full derivation is as follows:
\begin{align*}
 \int_1^2 \frac{(3x-1)(2x+3)}{x} \, dx &= \int_1^2 \frac{6x^2 + 7x -3}{x} \, dx \\
  &= \int_1^2 6x \, dx + \int_1^2 7 \, dx - \int_1^2 \frac{3}{x} \, dx \\
  &\phantom{=} \\
  &= 6 \left. \frac{x^2}{2} \right|_1^2 + 7 x \biggr \rvert_1^2 - 3 \ln x \biggr \rvert_1^2 \\
  &\phantom{=} \\
  &= 6\left( \frac{4}{2} - \frac{1}{2} \right) + 7(2-1) - 3 (\ln 2 - \ln 1) \\
  & \phantom{=} \\
  & = 16 - 3 \ln 2 \\
  & = 16 - \ln 2^3 \\
  & = 16 - \ln 8
\end{align*}
