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

\int \frac{1}{( x - 6 ) ( x - 3 ) ( x + 1 )} d x

y^{^{'}} = 100 x - x y

\frac{\partial}{\partial x} (- 9 x^{2} y^{2} - cos (y))

x^{3} + 8

\frac{1}{18 x ^{2}}, x, \frac{x}{8}

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