What is the integral from 0 to 4 of 1/(x^2+16) ?

The integral from 0 to 4 of 1/(x^2+16) is pi/(16)

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integral from 0 to 4 of 1/(x^2+16)

\int_{0}^{4} \frac{1}{x ^{2} + 16} d x

\frac{π}{16}

Decimal Notation

0.19634 \dots

Solution steps

\int_{0}^{4} \frac{1}{x ^{2} + 16} d x

Apply Integral Substitution

= \int_{0}^{1} \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_{0}^{1} \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)]_{0}^{1}

Compute the boundaries:

\frac{π}{4}

= \frac{1}{4} \cdot \frac{π}{4}

Simplify

= \frac{π}{16}

\frac{x}{4 + x}

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

\int 8 t e^{t} d t

4 \frac{d y}{d x} + 20 y = 5

\int \frac{2}{- x} d x

What is the integral from 0 to 4 of 1/(x^2+16) ?
The integral from 0 to 4 of 1/(x^2+16) is pi/(16)