What is the integral of (x^2)/(1+x^2) ?

The integral of (x^2)/(1+x^2) is -arctan(x)+x+C

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integral of (x^2)/(1+x^2)

\int \frac{x ^{2}}{1 + x ^{2}} d x

- arctan (x) + x + C

Solution steps

\int \frac{x ^{2}}{1 + x ^{2}} d x

\frac{x ^{2}}{1 + x ^{2}} = - \frac{1}{1 + x ^{2}} + 1

= \int - \frac{1}{1 + x ^{2}} + 1 d x

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= - \int \frac{1}{1 + x ^{2}} d x + \int 1 d x

\int \frac{1}{1 + x ^{2}} d x = arctan (x)

\int 1 d x = x

= - arctan (x) + x

Add a constant to the solution

= - arctan (x) + x + C

x y^{^{'}} + 4 y = 8 x^{4}, y (1) = 2

x^{2} + 3 x + 9, at x = 4

\frac{d}{d x} ((x + \frac{1}{x})^{2} - \frac{5}{3 x})

\frac{\partial}{\partial x} (x e^{3 y} sin (5 z))

n = 1 \sum \infty (\frac{5}{6})^{n}

What is the integral of (x^2)/(1+x^2) ?
The integral of (x^2)/(1+x^2) is -arctan(x)+x+C