$$\int^{3}_{1} \frac{x-4x^{2}}{x^{3}}\ \mathrm{d}x $$ I know the answer is $\frac23-4\ln3$ but I have no idea of how to find the antiderivative of the function. I'm studying for my final tomorrow and I would really appreciate it if anyone could please explain this problem to me step by step. Thank you!!
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$$\int^{3}_{1} \frac{x-4x^{2}}{x^{3}} \ \mathrm{d}x = \int^{3}_{1} \frac{1}{x^{2}} - \frac{4}{x} \ \mathrm{d}x = \left[-\frac{1}{x} - 4\ln{x}\right]^{3}_{1} = \left[-\frac{1}{3} - 4\ln{3} + 1\right] = \frac{2}{3} - 4\ln{3}$$
answered Apr 30, 2014 at 19:35
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$\begingroup$ Wait, can you explain it to me how you broke it up and what you did? It's still kind of confusing. $\endgroup$– DylanApr 30, 2014 at 19:36
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$\begingroup$ Sure. $\displaystyle \frac{x-4x^{2}}{x^{3}} = \frac{x}{x^{3}} - \frac{4x^{2}}{x^{3}}$ Now if we cancel since $x\not=0$: $\displaystyle \frac{x}{x^{3}} - \frac{4x^{2}}{x^{3}} = \frac{1}{x^{2}} - \frac{4}{x}$. $\endgroup$ Apr 30, 2014 at 19:38
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$\begingroup$ Got it. Now how do you find the anti derivative of a function like that? $\endgroup$– DylanApr 30, 2014 at 19:46
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$\begingroup$ For $\frac{1}{x^{2}}$ notice this is equal to $x^{-2}$ and you should know $$\int x^{a} \ \mathrm{d}x = \frac{x^{a+1}}{a+1} + C$$ for some $a\in\mathrm{R}$ (with $a=-2$). And for $\frac{4}{x}$ recall that $$\int \frac{1}{x} \ \mathrm{d}x = \ln{|x|} + C$$. $\endgroup$ Apr 30, 2014 at 19:50