What is the integral of 5/(9+4r^2) ?

The integral of 5/(9+4r^2) is 5/6 arctan(2/3 r)+C

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integral of 5/(9+4r^2)

\int \frac{5}{9 + 4 r ^{2}} d r

\frac{5}{6} arctan (\frac{2}{3} r) + C

Solution steps

\int \frac{5}{9 + 4 r ^{2}} d r

Take the constant out:

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

= 5 \cdot \int \frac{1}{9 + 4 r ^{2}} d r

Apply Integral Substitution

= 5 \cdot \int \frac{1}{6 ( u ^{2} + 1 )} d u

Take the constant out:

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

= 5 \cdot \frac{1}{6} \cdot \int \frac{1}{u ^{2} + 1} d u

Use the common integral:

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

= 5 \cdot \frac{1}{6} arctan (u)

Substitute back

u = \frac{2}{3} r

= 5 \cdot \frac{1}{6} arctan (\frac{2}{3} r)

Simplify

5 \cdot \frac{1}{6} arctan (\frac{2}{3} r) : \frac{5}{6} arctan (\frac{2}{3} r)

= \frac{5}{6} arctan (\frac{2}{3} r)

Add a constant to the solution

= \frac{5}{6} arctan (\frac{2}{3} r) + C

\frac{d}{d x} (e^{k x})

\int_{- 5}^{5} (25 - x^{2}) d x

\int \frac{1}{( 4 x + 5 ) ^{10}} d x

y^{^{''}} + 2 y^{^{'}} + 10 y = 0

f (x) = arcsin (5 x)

What is the integral of 5/(9+4r^2) ?
The integral of 5/(9+4r^2) is 5/6 arctan(2/3 r)+C