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

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

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

integral

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

\frac{1}{2} arctan (r) - \frac{r}{2 ( 1 + r ^{2} )} + C

Solution steps

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

Take the partial fraction of

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

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

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

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

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

Simplify

arctan (r) - \frac{1}{2} (arctan (r) + \frac{r}{1 + r ^{2}}) : \frac{1}{2} arctan (r) - \frac{r}{2 ( 1 + r ^{2} )}

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

Add a constant to the solution

= \frac{1}{2} arctan (r) - \frac{r}{2 ( 1 + r ^{2} )} + C

\frac{( 4 + 12 x )}{4}

(e^{x})^{- 2}

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

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

\frac{1}{( a ^{3} - 2 a )}

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