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This question already has an answer here:

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

$$u=x^2$$

$$du=2x \, dx$$

$$\int \frac{u}{u+1} \, du$$$$=\int \frac{u+1-1}{u+1}du$$$$=\int 1-\frac{1}{u+1} \, du$$

how should I continue? is there an algotherm for integrating rational functions?

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marked as duplicate by Andrés E. Caicedo, user147263, 6005, user296602, Shailesh Jan 4 '16 at 0:02

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

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    $\begingroup$ HINT: Use the substitution $v=1+u$ next $\endgroup$ – Mufasa Jan 3 '16 at 17:54
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    $\begingroup$ I really do not understand what your problem/question is. $\endgroup$ – Ron Gordon Jan 3 '16 at 17:54
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    $\begingroup$ You have come real close and then complete it as $u - ln(|u+1|)$. That is the reason why Ron is puzzled. $\endgroup$ – Satish Ramanathan Jan 3 '16 at 17:56
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    $\begingroup$ @Mufasa why should I use substitution? $\endgroup$ – gbox Jan 3 '16 at 17:59
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    $\begingroup$ You should just know that $\int\dfrac1{x+a}\,dx=\ln|x+a|$. It should be in your table. Either an allowed cheat sheet, or (preferrably) the table in your head. $\endgroup$ – Jyrki Lahtonen Jan 3 '16 at 18:01
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Notice, here is another simple method, $$\int \frac{2x^3}{x^2+1}\ dx$$ $$\int \frac{2x^3+2x-2x}{x^2+1}\ dx$$ $$=\int \frac{2x(x^2+1)-2x}{x^2+1}\ dx$$ $$=\int 2x\ dx-\int \frac{2x}{x^2+1}\ dx$$ $$=2\int x\ dx-\int \frac{d(x^2+1)}{x^2+1}$$ $$=\color{red}{x^2-\ln(x^2+1)+C}$$

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    $\begingroup$ Nice solution. Exactly how I would have proceeded. +1 $\endgroup$ – Mark Viola Jan 3 '16 at 18:16
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    $\begingroup$ Umpteenth time this is done on the site. Did you search for a duplicate? $\endgroup$ – Jyrki Lahtonen Jan 3 '16 at 18:24
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So far you have $\int 1 du - \int \frac{du}{u + 1}$.

Make the substitution $v = u + 1 \implies dv = du$ You get:

$u + c - \int \frac{dv}{v} = u - \ln\mid v \mid +\text{ } C$

Re substitute and you're done.

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No need of sub :

Just put $$2x^3=2x(1+x^2)-2x$$

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  • $\begingroup$ How would you continue? $\endgroup$ – zz20s Jan 3 '16 at 18:15

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