What is the integral of (20)/(1+x^2) ?

The integral of (20)/(1+x^2) is 20arctan(x)+C

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integral of (20)/(1+x^2)

\int \frac{20}{1 + x ^{2}} d x

20 arctan (x) + C

Solution steps

\int \frac{20}{1 + x ^{2}} d x

Take the constant out:

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

= 20 \cdot \int \frac{1}{1 + x ^{2}} d x

Use the common integral:

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

= 20 arctan (x)

Add a constant to the solution

= 20 arctan (x) + C

n = 1 \sum \infty \frac{1}{ln ( n )}

\frac{\partial}{\partial y} (arctan (x y^{2}))

\int 2 t e^{2 t} d t

x \to 0 - lim (\frac{3}{e ^{x} + 1})

\frac{\partial}{\partial y} ((\frac{x}{y}) e^{x y})

What is the integral of (20)/(1+x^2) ?
The integral of (20)/(1+x^2) is 20arctan(x)+C