$$\int\frac{2x^3}{x^2+1} \, dx$$
$$u=x^2$$
$$du=2x \, dx$$
$$\int \frac{u}{u+1} \, du$$$$=\int \frac{u+1-1}{u+1}du$$$$=\int 1-\frac{1}{u+1} \, du$$
how should I continue? is there an algotherm for integrating rational functions?
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3HINT: Use the substitution $v=1+u$ next – Mufasa Jan 03 '16 at 17:54
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1I really do not understand what your problem/question is. – Ron Gordon Jan 03 '16 at 17:54
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2You have come real close and then complete it as $u - ln(|u+1|)$. That is the reason why Ron is puzzled. – Satish Ramanathan Jan 03 '16 at 17:56
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1@Mufasa why should I use substitution? – gbox Jan 03 '16 at 17:59
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1You 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. – Jyrki Lahtonen Jan 03 '16 at 18:01
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1Because that will then simplify your integral of $$\int\frac{1}{u+1}du$$to:$$\int \frac{1}{v}dv$$which I assume you know how to do? – Mufasa Jan 03 '16 at 18:01
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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.
This Play Name
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Notice, here is another simple method, $$\int \frac{2x^3}{x^2+1}\ dx$$ $$\int \frac{2x^3+2x-2x}{x^2+1}\ dx$$ $$=\int \frac{2x(x^2+1)-2x}{x^2+1}\ dx$$ $$=\int 2x\ dx-\int \frac{2x}{x^2+1}\ dx$$ $$=2\int x\ dx-\int \frac{d(x^2+1)}{x^2+1}$$ $$=\color{red}{x^2-\ln(x^2+1)+C}$$
Harish Chandra Rajpoot
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3Umpteenth time this is done on the site. Did you search for a duplicate? – Jyrki Lahtonen Jan 03 '16 at 18:24
