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

Question

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

Answer 1

That's the most straightforward way.

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

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

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Added (to address comment below):
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 $x$.

Answer 2

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

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

Answer 3

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

Answer 4

$\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 : $\ln(x^2+1)+C$