What is the integral of 9/(64+4r^2) ?

The integral of 9/(64+4r^2) is 9/16 arctan(r/4)+C

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integral of 9/(64+4r^2)

\int \frac{9}{64 + 4 r ^{2}} d r

\frac{9}{16} arctan (\frac{r}{4}) + C

Solution steps

\int \frac{9}{64 + 4 r ^{2}} d r

Take the constant out:

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

= 9 \cdot \int \frac{1}{64 + 4 r ^{2}} d r

Apply Integral Substitution

= 9 \cdot \int \frac{1}{16 ( u ^{2} + 1 )} d u

Take the constant out:

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

= 9 \cdot \frac{1}{16} \cdot \int \frac{1}{u ^{2} + 1} d u

Use the common integral:

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

= 9 \cdot \frac{1}{16} arctan (u)

Substitute back

u = \frac{2}{8} r

= 9 \cdot \frac{1}{16} arctan (\frac{2}{8} r)

Simplify

9 \cdot \frac{1}{16} arctan (\frac{2}{8} r) : \frac{9}{16} arctan (\frac{r}{4})

= \frac{9}{16} arctan (\frac{r}{4})

Add a constant to the solution

= \frac{9}{16} arctan (\frac{r}{4}) + C

\frac{d}{d x} (9 x^{3} - 7 x)

\frac{d}{d x} ((x^{3} + 3)^{5})

\int \frac{1}{( x + 4 ) x ^{2} + 8 x + 12} d x

\frac{d}{d x} (\frac{x ^{4}}{2})

\int_{- 1}^{4} (x + 3) d x

What is the integral of 9/(64+4r^2) ?
The integral of 9/(64+4r^2) is 9/16 arctan(r/4)+C