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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

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  • $\begingroup$ Open brackets and solve directly. Remember that (ln(x))' = 1/x. $\endgroup$ – hvedrung Aug 13 '15 at 12:46
  • $\begingroup$ can someone tell me what code I need to put in so the x comes underneath? $\endgroup$ – joe Aug 13 '15 at 12:51
  • $\begingroup$ @joe I can't edit your post to fix the latex, but you can look at the code on my answer for how all of that is done. Or just to get fractions in general, you want \frac{numerator}{denominator}. So, here you type: \int_1^2 \frac{(3x-1)(2x+3)}{x}. with dollar signs, of course. $\endgroup$ – Bamboo Aug 13 '15 at 12:53
  • $\begingroup$ It's done now, I get what you mean mate it's just a bit confusing lol $\endgroup$ – joe Aug 13 '15 at 12:55
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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*}

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$$\int_{1}^{2}\frac{6x^2+7x-3}{x}dx$$ $$=\int_{1}^{2}\left(6x+7-\frac{3}{x}\right)dx$$ $$=\left(3x^2+7x-3\ln |x|\right)_{1}^{2}$$ $$=\left(3(2)^2+7(2)-3\ln |2|-(3(1)^2+7(1)-3\ln |1|)\right)$$ $$=16-3\ln 2$$ $$=16-\ln 2^3=16-\ln 8$$

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expanding the integrad and dividing by $x$ we get $$\int_{1}^{2}7+6x-\frac{3}{x}dx$$ the indefinite integral is given by $$7x+3x^2-3\ln|x|+C$$

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$\displaystyle \int_{1}^{2}\frac{(3x-1)\cdot (2x+3)}{x}dx = \int_{1}^{2}\frac{6x^2+7x-3}{x}dx = \int_{1}^{2}6xdx+\int_{1}^{2}7dx-\int_{1}^{2}\frac{1}{x}dx$

$\displaystyle = [3x^2]_{1}^{2}+[7x]_{1}^{2}-[3\ln|x|]_{1}^{2} = 16-3\ln(2)$

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