What is the integral from 0 to infinity of 5/(x^2+1) ?

The integral from 0 to infinity of 5/(x^2+1) is (5pi)/2

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integral from 0 to infinity of 5/(x^2+1)

\int_{0}^{\infty} \frac{5}{x ^{2} + 1} d x

\frac{5 π}{2}

Decimal Notation

7.85398 \dots

Solution steps

\int_{0}^{\infty} \frac{5}{x ^{2} + 1} d x

Compute the indefinite integral:

\int \frac{5}{x ^{2} + 1} d x = 5 arctan (x) + C

Compute the boundaries:

\int_{0}^{\infty} \frac{5}{x ^{2} + 1} d x = 5 \cdot \frac{π}{2} - 0

= 5 \cdot \frac{π}{2} - 0

Simplify

= \frac{5 π}{2}

\int \frac{1}{x - 8} d x

n = 1 \sum \infty (cos (29))^{n}

e^{\frac{2}{x}}

\frac{d ^{2}}{d x ^{2}} (\frac{1}{3 + x})

f (x) = (x^{2} + 3)^{4}

What is the integral from 0 to infinity of 5/(x^2+1) ?
The integral from 0 to infinity of 5/(x^2+1) is (5pi)/2