How do you integrate $\frac{x^{2}}{x^{2} + 1}$ $x^2/(x^2+1)$ ?

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Answer 1 · 2016-06-28T07:40:29

$x - arctan x + C$ $x - arctan x + C$

$\frac{x^{2}}{x^{2} + 1} = \frac{x^{2} + 1 - 1}{x^{2} + 1} = 1 - \frac{1}{x^{2} + 1}$ $x^2/(x^2+1) = (x^2+1 - 1)/(x^2+1) = 1 - ( 1)/(x^2+1)$

$int \ 1 - ( 1)/(x^2+1) \ dx$

$= x - color(red)(int \ ( 1)/(x^2+1) \ dx )$

in terms of the red bit, use sub $x = tan t, dx = sec^2 t \ dt$

this makes it

$\int \ ( 1)/(tan^2 t+1) \ sec^2 t \ dt$

$= \int \ ( 1)/(sec^2 t) \ sec^2 t \ dt$

$= \int \ dt$

$= arctan x - C$

So the full integral is

$x - arctan x + C$

How do you integrate x^2/(x^2+1)x2x2+1x^2/(x^2+1)?