What is the integral of (13)/(r^2+1) ?

The integral of (13)/(r^2+1) is 13arctan(r)+C

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integral of (13)/(r^2+1)

\int \frac{13}{r ^{2} + 1} d r

13 arctan (r) + C

Solution steps

\int \frac{13}{r ^{2} + 1} d r

Take the constant out:

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

= 13 \cdot \int \frac{1}{r ^{2} + 1} d r

Use the common integral:

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

= 13 arctan (r)

Add a constant to the solution

= 13 arctan (r) + C

e^{x} y \frac{d y}{d x} = e^{- y} + e^{- 8 x - y}

\frac{\partial}{\partial x} ((5 y^{4}) cos (5 x))

f (x) = x + 5, (11, 4)

h (t) = (t + 2)^{\frac{2}{3}} (2 t^{2} - 3)^{3}

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

What is the integral of (13)/(r^2+1) ?
The integral of (13)/(r^2+1) is 13arctan(r)+C