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

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

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

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

- arctan (x) + x + C

Solution steps

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

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

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

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

\int 1 d x = x

= - arctan (x) + x

Add a constant to the solution

= - arctan (x) + x + C

(y^{4} x) \frac{d y}{d x} = 1 + x, y (1) = 2

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

\frac{d}{d y} (36 - y^{2})

\frac{\partial}{\partial x} (7 y e^{8 x})

x \to 0 lim (\frac{3}{e ^{x} - 1} - \frac{1}{3})

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