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

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

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

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

\frac{π}{2}

Decimal Notation

1.57079 \dots

Solution steps

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

Take the constant out:

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

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

Use the common integral:

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

= 2 [arctan (t)]_{0}^{1}

Compute the boundaries:

\frac{π}{4}

= 2 \cdot \frac{π}{4}

Simplify

= \frac{π}{2}

\frac{e ^{x}}{4 x ^{2} + 3}

f (x) = 2 π r^{2} + \frac{1000}{r}

y = \frac{6}{x}, (36, \frac{6}{6})

\frac{d y}{d x} = - 2 y + e^{x}

\int_{0}^{\frac{π}{6}} 4 cos^{5} (3 x) d x

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