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

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

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integral from 1 to infinity of 9/(x^2+1)

\int_{1}^{\infty} \frac{9}{x ^{2} + 1} d x

\frac{9 π}{4}

Decimal Notation

7.06858 \dots

Solution steps

\int_{1}^{\infty} \frac{9}{x ^{2} + 1} d x

Compute the indefinite integral:

\int \frac{9}{x ^{2} + 1} d x = 9 arctan (x) + C

Compute the boundaries:

\int_{1}^{\infty} \frac{9}{x ^{2} + 1} d x = 9 \cdot \frac{π}{2} - 9 \cdot \frac{π}{4}

= 9 \cdot \frac{π}{2} - 9 \cdot \frac{π}{4}

Simplify

= \frac{9 π}{4}

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