What is the integral from 0 to 1 of 9/(1+t^2) ?

The integral from 0 to 1 of 9/(1+t^2) is (9pi)/4

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integral from 0 to 1 of 9/(1+t^2)

\int_{0}^{1} \frac{9}{1 + t ^{2}} d t

\frac{9 π}{4}

Decimal Notation

7.06858 \dots

Solution steps

\int_{0}^{1} \frac{9}{1 + t ^{2}} d t

Take the constant out:

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

= 9 \cdot \int_{0}^{1} \frac{1}{1 + t ^{2}} d t

Use the common integral:

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

= 9 [arctan (t)]_{0}^{1}

Compute the boundaries:

\frac{π}{4}

= 9 \cdot \frac{π}{4}

Simplify

= \frac{9 π}{4}

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What is the integral from 0 to 1 of 9/(1+t^2) ?
The integral from 0 to 1 of 9/(1+t^2) is (9pi)/4