What is the integral from 0 to infinity of 9/(x^2+4) ?

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

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integral from 0 to infinity of 9/(x^2+4)

\int_{0}^{\infty} \frac{9}{x ^{2} + 4} d x

\frac{9 π}{4}

Decimal

7.06858 \dots

Solution steps

\int_{0}^{\infty} \frac{9}{x ^{2} + 4} d x

Compute the indefinite integral:

\int \frac{9}{x ^{2} + 4} d x = \frac{9}{2} arctan (\frac{x}{2}) + C

Compute the boundaries:

\int_{0}^{\infty} \frac{9}{x ^{2} + 4} d x = \frac{9 π}{4} - 0

= \frac{9 π}{4} - 0

Simplify

= \frac{9 π}{4}

(4, 2), (- 16, - 10)

t^{- \frac{1}{2}}

\int \frac{x ^{4} - 2 x ^{2} + 2}{( x ^{2} - 2 x + 2 ) ^{2}} d x

\int \frac{e ^{1 + l n (x)}}{x} d x

y^{^{'}} = \frac{y ^{2} + 1}{x}

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