How to integrate $\large \frac{2x}{1+x^{2}}$

How to integrate $\;\large \frac{2x}{1+x^{2}}\;?\;$ Do I need to use u-substitution for $\,(1+x^2)\,$?

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That’s certainly the easiest way: it makes the integral a very simple one. – Brian M. Scott Feb 3 '13 at 21:57
It's much easier to judge whether an approach to a problem is useful after you have tried it out rather than before. – Hurkyl Feb 4 '13 at 16:34

That's the most straightforward way.

Let $u = 1 + x^2,\;$ then $du = 2x \,dx.\;$ Now you're all set to substitute:

$$\int \dfrac{2x\,dx}{1+x^2} \;\;=\; \;\int \dfrac{du}{u}\;\;=\;\; \ln |u| + C \;= \;\ln(1 + x^2) + C$$

Note that we can omit the "absolute value sign" surrounding $\,(1 + x^2)\,$ when substituting $\,u = 1+x^2\,$ in the final expression because we know that $\,(1 + x^2) > 0\,$ for all real $x$.

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One small point, $u=|u|$ in this case. Otherwise we would have to write $ln|1+x^2|+C$ – Baby Dragon Feb 3 '13 at 23:27
Note that $1 + x^2 >0$ for all values of x. That's the only reason I dropped them, as we are guaranteed that $|1 + x^2| = 1 + x^2$ – amWhy Feb 3 '13 at 23:55

$$2x=(1+x^2)'$$

$$\int{\frac{(1+x^2)'}{(1+x^2)}}dx$$

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thanks for suggestion – Iuli Feb 3 '13 at 22:01
Looks good now; +1. – Brian M. Scott Feb 3 '13 at 22:12

$\textbf{Hint:}$ You don't need any fancy technique. Look at an antiderivative table and try to work it out.

Scroll over the grey area for the solution.

Note that $\displaystyle \int \frac{2x}{1+x^2}dx=\int \frac{(1+x^2)'}{1+x^2}dx=\log {\bigl(|1+x^2|\bigr)}=\log {(1+x^2)}$

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The solution is not an integrand. You have the solution posted as an integrand. The point was to integrate the function to the left, which is the integrand. – amWhy Feb 3 '13 at 22:18
@amWhy That was a funny mistake. Thanks. – Git Gud Feb 3 '13 at 22:21

$\displaystyle\frac{\mathrm{d}}{\mathrm{d}x} \ln u(x) =\frac{u'(x)}{u(x)}$

in this case : $u(x)=x^2+1$ so the primitive is : $\ln(x^2+1)+C$

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