What is the integral from 0 to 1 of 8/(t^2+1) ?

The integral from 0 to 1 of 8/(t^2+1) is 2pi

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integral from 0 to 1 of 8/(t^2+1)

\int_{0}^{1} \frac{8}{t ^{2} + 1} d t

2 π

Decimal

6.28318 \dots

Solution steps

\int_{0}^{1} \frac{8}{t ^{2} + 1} d t

Take the constant out:

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

= 8 \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)

= 8 [arctan (t)]_{0}^{1}

Compute the boundaries:

\frac{π}{4}

= 8 \cdot \frac{π}{4}

Simplify

= 2 π

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What is the integral from 0 to 1 of 8/(t^2+1) ?
The integral from 0 to 1 of 8/(t^2+1) is 2pi