What is the integral of (16)/(1+x^2) ?

The integral of (16)/(1+x^2) is 16arctan(x)+C

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integral of (16)/(1+x^2)

\int \frac{16}{1 + x ^{2}} d x

16 arctan (x) + C

Solution steps

\int \frac{16}{1 + x ^{2}} d x

Take the constant out:

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

= 16 \cdot \int \frac{1}{1 + x ^{2}} d x

Use the common integral:

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

= 16 arctan (x)

Add a constant to the solution

= 16 arctan (x) + C

\frac{d y}{d x} = 2 (1 - y)

\frac{\partial}{\partial y} (\frac{- 1}{x ^{2} + y ^{2}})

\frac{\partial}{\partial x} (e^{(4 x y)})

y = x lo g_{e} (x)

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

What is the integral of (16)/(1+x^2) ?
The integral of (16)/(1+x^2) is 16arctan(x)+C