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

The integral of 1/((x^2+16)) is 1/4 arctan(x/4)+C

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

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

\frac{1}{4} arctan (\frac{x}{4}) + C

Solution steps

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

Apply Integral Substitution

= \int \frac{1}{4 ( u ^{2} + 1 )} d u

Take the constant out:

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

= \frac{1}{4} \cdot \int \frac{1}{u ^{2} + 1} d u

Use the common integral:

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

= \frac{1}{4} arctan (u)

Substitute back

u = \frac{x}{4}

= \frac{1}{4} arctan (\frac{x}{4})

Add a constant to the solution

= \frac{1}{4} arctan (\frac{x}{4}) + C

\int_{0}^{x} t^{4} + t^{6} d t

y^{^{''}} - 2 y^{^{'}} - 2 y = 0

x \to n - lim ([x])

\frac{d}{d x} ((3 - e^{x})^{8})

\frac{d}{d x} (180 - \frac{x ^{2}}{45})

What is the integral of 1/((x^2+16)) ?
The integral of 1/((x^2+16)) is 1/4 arctan(x/4)+C