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

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

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

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

2 π

Decimal Notation

6.28318 \dots

Solution steps

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

= 4 π - 2 π

Simplify

= 2 π

\int (csc^{2} (x) - 2 e^{x}) d x

x \to 2 lim (x^{3} + 4 x^{2} - 2 x + 2)

f (x) = 2 x^{2} e^{x} (x^{2} - 3 x + 1)

\int 2 x \cdot e^{x} d x

\int t^{8} d t

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