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How to solve next integral: $$\int\frac{x^3+4x^2-5}{x^2}dx$$ I am using power rule for top part wich produce this: $$\int\frac{\frac{x^2}{2}+4x-5x}{x^2}dx$$ Does my calculation right? How can I continue from here? Please describe all steps and rules used for solving this integral.

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closed as off-topic by Jack D'Aurizio, Namaste, Arnaldo, Leucippus, Daniel W. Farlow Jun 7 '17 at 0:49

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    $\begingroup$ It is obviously wrong. Otherwise $\frac{x^2}{2}=\int x\,dx = \int\frac{x^2}{x}\,dx = \frac{\frac{x^3}{3}}{x}=\frac{x^2}{3}$, according to your logic. $\endgroup$ – Jack D'Aurizio Jun 6 '17 at 20:18
  • $\begingroup$ Which power rule? $\endgroup$ – Qwerty Jun 6 '17 at 20:18
  • $\begingroup$ I can not see any rule. $x^3\to \frac12x^2$, $x^2\to x$ and $1\to x$? That can not be right. $\endgroup$ – Mundron Schmidt Jun 6 '17 at 20:21
  • $\begingroup$ remember to simplify the exponents first!! $\endgroup$ – Saketh Malyala Jun 6 '17 at 20:32
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What you already have is incorrect. Try this:

$$ \int \frac{x^3+4x^2-5}{x^2}dx = \int \frac{x^3}{x^2} + \frac{4x^2}{x^2} -\frac{5}{x^2} dx $$

$$ =\int (x) dx + \int 4 dx + \int \frac{5}{x^2} dx $$

$$ =\frac{x^2}{2} + 4x + \frac{5}{x} + C $$

Overview of Integration rules:

$$ \int x^n \hspace{2 mm}dx= \frac{x^{n+1}}{n+1} + C $$

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  • $\begingroup$ How $\frac{x^3}{x^2}$ becomes $\frac{x^2}{2}$? and how $-\frac{5}{x^2}$ becomes ${5}{x}$? $\endgroup$ – IntoTheDeep Jun 6 '17 at 20:25
  • $\begingroup$ $$ \int x^n = \frac{x^{n+1}}{n+1} + C $$ Check this out as a reference: mathsisfun.com/calculus/integration-rules.html $\endgroup$ – Dashi Jun 6 '17 at 20:30
  • $\begingroup$ Does this reference apply for $-\frac{5}{x^2}$? $\endgroup$ – IntoTheDeep Jun 6 '17 at 20:34
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    $\begingroup$ Yes: $$ \int \frac{-5}{x^2} dx= -5 \int x^{-2} dx = -5 \left( \frac{x^{-2+1}}{-2+1} \right)+C = -5 (-x^{-1}) + C = \frac{5}{x} + C $$ $\endgroup$ – Dashi Jun 6 '17 at 20:36
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    $\begingroup$ $$ \int \frac{x^3}{x^2} dx = \int x^1 \hspace{2 mm} dx = \frac{x^{1+1}}{1+1} + C = \frac{x^2}{2} + C $$ $\endgroup$ – Dashi Jun 6 '17 at 20:51
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Hint:

$$\int \frac{x^3+4x^2-5}{x^2}dx=\int x+4-5x^{-2}dx$$

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  • $\begingroup$ Wrong and no steps added. Please read question before answer $\endgroup$ – IntoTheDeep Jun 6 '17 at 20:23
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    $\begingroup$ @TeodorKolev Wrong ? What is your reasoning ? $\endgroup$ – callculus Jun 6 '17 at 20:23
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What you want to use is the linearity of the integral: \begin{equation} \int \frac{x^3 + 4x^2 -5}{x^2} dx = \int \frac{x^3}{x^2} dx + \int \frac{ 4x^2}{x^2} dx - \int \frac{5}{x^2} dx.\end{equation}Can you continue from here on out?

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  • $\begingroup$ No, I can't. Sorry $\endgroup$ – IntoTheDeep Jun 6 '17 at 20:28
  • $\begingroup$ Not a problem, we are all here to learn! I think numerous people have answered already so I'll leave it with this. In the future you it would help you to specify how much detail you need, that makes it easier for people to help you. $\endgroup$ – M.v.Roozendaal Jun 6 '17 at 21:10
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Break down the sum and reduce the polynomial then use the linearity of the integral to get

$$\begin{align}\int\frac{x^3+4x^2-5}{x^2}dx=&\int{\left(x+4-{5\over x^2}\right)}dx\\ =&\int{xdx}+4\int{dx}-5\int\frac{dx}{x^2}\\ =&{x^2\over 2}+4x+{5\over x}+C\end{align}$$

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    $\begingroup$ No steps are added. I do not understand anything $\endgroup$ – IntoTheDeep Jun 6 '17 at 20:22
  • $\begingroup$ It should be better now $\endgroup$ – marwalix Jun 6 '17 at 20:26

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