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

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

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

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

\frac{9 π}{2}

Decimal Notation

14.13716 \dots

Solution steps

\int_{0}^{\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_{0}^{\infty} \frac{9}{x ^{2} + 1} d x = 9 \cdot \frac{π}{2} - 0

= 9 \cdot \frac{π}{2} - 0

Simplify

= \frac{9 π}{2}

x \to 0 lim (\frac{x}{x + sin ( x )})

n = 1 \sum \infty (- \frac{1}{3})^{n}

x \to 2 lim (8 + x^{3})

\int \frac{2 y ^{2} - 16}{y ^{2} + 5 y + 6} d y

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