What is the integral of 7/(16+9r^2) ?

The integral of 7/(16+9r^2) is 7/12 arctan(3/4 r)+C

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integral of 7/(16+9r^2)

\int \frac{7}{16 + 9 r ^{2}} d r

\frac{7}{12} arctan (\frac{3}{4} r) + C

Solution steps

\int \frac{7}{16 + 9 r ^{2}} d r

Take the constant out:

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

= 7 \cdot \int \frac{1}{16 + 9 r ^{2}} d r

Apply Integral Substitution

= 7 \cdot \int \frac{1}{12 ( u ^{2} + 1 )} d u

Take the constant out:

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

= 7 \cdot \frac{1}{12} \cdot \int \frac{1}{u ^{2} + 1} d u

Use the common integral:

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

= 7 \cdot \frac{1}{12} arctan (u)

Substitute back

u = \frac{3}{4} r

= 7 \cdot \frac{1}{12} arctan (\frac{3}{4} r)

Simplify

7 \cdot \frac{1}{12} arctan (\frac{3}{4} r) : \frac{7}{12} arctan (\frac{3}{4} r)

= \frac{7}{12} arctan (\frac{3}{4} r)

Add a constant to the solution

= \frac{7}{12} arctan (\frac{3}{4} r) + C

\frac{d}{d x} ((4 x)^{3})

y = \frac{7}{x}, (1, 7)

h (t) = (t^{4} - 1)^{3} (t^{3} + 1)^{4}

\frac{d}{d x} ((6 x + 8) (4 x + 9)^{5})

\frac{d}{d x} (x e^{- \frac{1}{2} x})

What is the integral of 7/(16+9r^2) ?
The integral of 7/(16+9r^2) is 7/12 arctan(3/4 r)+C