What is the integral from 0 to infinity of 8/(x^2+1) ?

The integral from 0 to infinity of 8/(x^2+1) is 4pi

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integral from 0 to infinity of 8/(x^2+1)

\int_{0}^{\infty} \frac{8}{x ^{2} + 1} d x

4 π

Decimal Notation

12.56637 \dots

Solution steps

\int_{0}^{\infty} \frac{8}{x ^{2} + 1} d x

Compute the indefinite integral:

\int \frac{8}{x ^{2} + 1} d x = 8 arctan (x) + C

Compute the boundaries:

\int_{0}^{\infty} \frac{8}{x ^{2} + 1} d x = 4 π - 0

= 4 π - 0

Simplify

= 4 π

sec^{2} (4 x)

\frac{\partial}{\partial x} (x^{\frac{8}{x}})

\int \frac{12 x ^{3} - 1}{3 x ^{4} - x + 10} d x

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

\int \frac{( x - 1 ) ^{2}}{x ^{2}} d x

What is the integral from 0 to infinity of 8/(x^2+1) ?
The integral from 0 to infinity of 8/(x^2+1) is 4pi