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

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

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

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

\frac{π}{16}

Decimal Notation

0.19634 \dots

Solution steps

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

Apply Integral Substitution

= \int_{0}^{1} \frac{1}{4 ( 1 + u ^{2} )} 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}{1 + u ^{2}} d u

Use the common integral:

\int \frac{1}{1 + u ^{2}} 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{d y}{d t} = y^{2} t^{2}

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

y^{^{''}} + 46 y^{^{'}} + 24 y = 0

\frac{d}{d x} (cot (x^{3}))

x \to 0 lim (+ (\frac{1}{x}))

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