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

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

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

integral

\frac{r ^{2}}{( 1 + r ^{2} )}

- arctan (r) + r + C

Solution steps

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

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

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

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 + r ^{2}} d r + \int 1 d r

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

\int 1 d r = r

= - arctan (r) + r

Add a constant to the solution

= - arctan (r) + r + C

\frac{1 - sin ^{2} ( x )}{sin ^{2} ( x )}

\frac{1}{( 3 x ^{2} - 4 x )}

0.7 5^{x}

1 + y^{3}

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

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