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

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

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

integral

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

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

Solution steps

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

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

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

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

= \frac{1}{d} \cdot \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

= \frac{1}{d} (- \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

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

Add a constant to the solution

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

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

\frac{1}{( 5 + 2 x + x ^{2} ) ^{\frac{3}{2}}}

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