# Integrating $\int \frac{Ax^2}{B+x^2}dx$

I was hoping someone could help explain to me how to integrate $$\int \frac{Ax^2}{B+x^2}dx$$

I plugged it in on wolfram and got the correct answer of $A(x-\sqrt{B}\arctan(\frac{x}{\sqrt{B}}))$.

I am just scared that something like this might show up on our final and I don't really get how it was done.

Any help is really appreciated.

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## 2 Answers

Just use the following transformations: $$\frac{x^2}{B+x^2}=\frac{B-B+x^2}{B+x^2}=1-\frac{B}{B+x^2}=1-\frac{1}{\left(\frac{x}{\sqrt{B}}\right)^2+1}$$ So $$\int \frac{Ax^2}{B+x^2}dx=A \int \frac{x^2}{B+x^2}dx=A\int dx - A\int\frac{1}{\left(\frac{x}{\sqrt{B}}\right)^2+1} dx=A(x-\sqrt{B}\arctan(\frac{x}{\sqrt{B}}))$$

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that's it.simple and clear –  masmoudihoussem Mar 25 at 17:23
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$$\int \frac{Ax^2}{B+x^2}dx=A\int\frac{x^2+B-B}{B+x^2}dx=A\int dx-AB\int\frac{dx}{B+x^2}$$ and $$\int\frac{dx}{B+x^2}=\frac{\sqrt{B}}{B}\int\frac{dx/\sqrt{B}}{1+(x/\sqrt{B})^2}=\frac{1}{\sqrt{B}}\int\frac{dv}{1+v^2}$$

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Thank you for the explanation –  user68203 Mar 26 at 0:46
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