What is the integral of (21)/(4+25r^2) ?

The integral of (21)/(4+25r^2) is 21/10 arctan(5/2 r)+C

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integral of (21)/(4+25r^2)

\int \frac{21}{4 + 25 r ^{2}} d r

\frac{21}{10} arctan (\frac{5}{2} r) + C

Solution steps

\int \frac{21}{4 + 25 r ^{2}} d r

Take the constant out:

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

= 21 \cdot \int \frac{1}{4 + 25 r ^{2}} d r

Apply Integral Substitution

= 21 \cdot \int \frac{1}{10 ( u ^{2} + 1 )} d u

Take the constant out:

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

= 21 \cdot \frac{1}{10} \cdot \int \frac{1}{u ^{2} + 1} d u

Use the common integral:

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

= 21 \cdot \frac{1}{10} arctan (u)

Substitute back

u = \frac{5}{2} r

= 21 \cdot \frac{1}{10} arctan (\frac{5}{2} r)

Simplify

21 \cdot \frac{1}{10} arctan (\frac{5}{2} r) : \frac{21}{10} arctan (\frac{5}{2} r)

= \frac{21}{10} arctan (\frac{5}{2} r)

Add a constant to the solution

= \frac{21}{10} arctan (\frac{5}{2} r) + C

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

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

\frac{\partial}{\partial x} (6 x^{2} y + 6 x y^{3})

x \to - 5 lim (g (x))

\frac{d}{d x} (\frac{x}{x ^{2} + y ^{2} + z ^{2}})

What is the integral of (21)/(4+25r^2) ?
The integral of (21)/(4+25r^2) is 21/10 arctan(5/2 r)+C