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

The integral from 1 to infinity of 5/(x^2+1) is (5pi)/4

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

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

\frac{5 π}{4}

Decimal Notation

3.92699 \dots

Solution steps

\int_{1}^{\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_{1}^{\infty} \frac{5}{x ^{2} + 1} d x = 5 \cdot \frac{π}{2} - 5 \cdot \frac{π}{4}

= 5 \cdot \frac{π}{2} - 5 \cdot \frac{π}{4}

Simplify

= \frac{5 π}{4}

n = 2 \sum \infty \frac{( - 1 ) ^{n}}{n}

\int x^{3} e^{- 2 x} d x

\frac{d}{d j} (12 i + 15 j)

f (x) = 2 x^{2} + 5 x - 1, at x = - 3

\frac{d}{d x} ((4 x^{2} + 5) (5 x + 2)^{4})

What is the integral from 1 to infinity of 5/(x^2+1) ?
The integral from 1 to infinity of 5/(x^2+1) is (5pi)/4