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

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

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integral from 0 to 1 of 1/(t^2+1)

\int_{0}^{1} \frac{1}{t ^{2} + 1} d t

\frac{π}{4}

Decimal Notation

0.78539 \dots

Solution steps

\int_{0}^{1} \frac{1}{t ^{2} + 1} d t

Use the common integral:

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

= [arctan (t)]_{0}^{1}

Compute the boundaries:

\frac{π}{4}

= \frac{π}{4}

\frac{2 x + 2 y - 3}{- x - y - 1} = \frac{d y}{d x}

x^{2} y^{^{''}} - 2 x y^{^{'}} + 2 y = x^{3} e^{x}

\int \frac{5 x}{9 x ^{2} - 4} d x

n = 0 \sum \infty e^{\frac{- n}{2}}

(x + y) d x + (x) d y = 0

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