What is the integral from-infinity to infinity of 1/(2x^2-8x+40) ?

The integral from-infinity to infinity of 1/(2x^2-8x+40) is pi/8

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integral from-infinity to infinity of 1/(2x^2-8x+40)

\int_{- \infty}^{\infty} \frac{1}{2 x ^{2} - 8 x + 40} d x

\frac{π}{8}

Decimal Notation

0.39269 \dots

Solution steps

\int_{- \infty}^{\infty} \frac{1}{2 x ^{2} - 8 x + 40} d x

Compute the indefinite integral:

\int \frac{1}{2 x ^{2} - 8 x + 40} d x = \frac{1}{8} arctan (\frac{x - 2}{4}) + C

Compute the boundaries:

\int_{- \infty}^{\infty} \frac{1}{2 x ^{2} - 8 x + 40} d x = \frac{π}{16} - (- \frac{π}{16})

= \frac{π}{16} - (- \frac{π}{16})

Simplify

= \frac{π}{8}

\int_{0}^{2} (x^{2} + 1)^{2} d x

\int_{0}^{1} e^{- x} d x

\int_{0}^{4} x + 2 d x

\int_{2}^{4} (- x^{2} + 2 x + 8) d x

\int_{0}^{1} (x^{12} + 1 2^{x}) d x

What is the integral from-infinity to infinity of 1/(2x^2-8x+40) ?
The integral from-infinity to infinity of 1/(2x^2-8x+40) is pi/8