What is the integral of (16)/(25+9r^2) ?

The integral of (16)/(25+9r^2) is 16/15 arctan(3/5 r)+C

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integral of (16)/(25+9r^2)

\int \frac{16}{25 + 9 r ^{2}} d r

\frac{16}{15} arctan (\frac{3}{5} r) + C

Solution steps

\int \frac{16}{25 + 9 r ^{2}} d r

Take the constant out:

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

= 16 \cdot \int \frac{1}{25 + 9 r ^{2}} d r

Apply Integral Substitution

= 16 \cdot \int \frac{1}{15 ( u ^{2} + 1 )} d u

Take the constant out:

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

= 16 \cdot \frac{1}{15} \cdot \int \frac{1}{u ^{2} + 1} d u

Use the common integral:

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

= 16 \cdot \frac{1}{15} arctan (u)

Substitute back

u = \frac{3}{5} r

= 16 \cdot \frac{1}{15} arctan (\frac{3}{5} r)

Simplify

16 \cdot \frac{1}{15} arctan (\frac{3}{5} r) : \frac{16}{15} arctan (\frac{3}{5} r)

= \frac{16}{15} arctan (\frac{3}{5} r)

Add a constant to the solution

= \frac{16}{15} arctan (\frac{3}{5} r) + C

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What is the integral of (16)/(25+9r^2) ?
The integral of (16)/(25+9r^2) is 16/15 arctan(3/5 r)+C