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

\frac{d y}{d t} + y = 2

\int \frac{6 x + 6}{5 x ^{2} + 26 x + 5} d x

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

\frac{\partial}{\partial x} (sin (x^{3} + y^{2}))

y^{^{'}} + 5 y = 1

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