What is the integral of (10)/(r^2+1) ?

The integral of (10)/(r^2+1) is 10arctan(r)+C

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integral of (10)/(r^2+1)

\int \frac{10}{r ^{2} + 1} d r

10 arctan (r) + C

Solution steps

\int \frac{10}{r ^{2} + 1} d r

Take the constant out:

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

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

Use the common integral:

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

= 10 arctan (r)

Add a constant to the solution

= 10 arctan (r) + C

y = 8 - 8 x^{2}, y = 0

6 x^{9} e^{x}

\frac{d}{d x} ((1 - x - x^{3}) (1 - x + x^{3}))

\frac{\partial}{\partial x} (ln (y - x) - \frac{x}{y - x})

\frac{d}{d x} (40320 x)

What is the integral of (10)/(r^2+1) ?
The integral of (10)/(r^2+1) is 10arctan(r)+C