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

The integral of 9/(16+25r^2) is 9/20 arctan(5/4 r)+C

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

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

\frac{9}{20} arctan (\frac{5}{4} r) + C

Solution steps

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

Apply Integral Substitution

= 9 \cdot \int \frac{1}{20 ( 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}{20} \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}{20} arctan (u)

Substitute back

u = \frac{5}{4} r

= 9 \cdot \frac{1}{20} arctan (\frac{5}{4} r)

Simplify

9 \cdot \frac{1}{20} arctan (\frac{5}{4} r) : \frac{9}{20} arctan (\frac{5}{4} r)

= \frac{9}{20} arctan (\frac{5}{4} r)

Add a constant to the solution

= \frac{9}{20} arctan (\frac{5}{4} r) + C

\frac{d}{d x} ((x + 2) (3 x^{2} - 4)^{2})

\frac{d}{d x} (cos (3 x) 3)

y^{^{''}} - 8 y^{^{'}} + 9 y = x e^{x}

l a pl a ce t r an s f or m sin (2 (t - \frac{π}{2}))

a re a x^{2} - 4 x, x - 4

What is the integral of 9/(16+25r^2) ?
The integral of 9/(16+25r^2) is 9/20 arctan(5/4 r)+C