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

The integral from 5 to infinity of 1/(x^2-4) is (ln(7/2)-ln(3/2))/4

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integral from 5 to infinity of 1/(x^2-4)

\int_{5}^{\infty} \frac{1}{x ^{2} - 4} d x

\frac{ln ( \frac{7}{2} ) - ln ( \frac{3}{2} )}{4}

Decimal Notation

0.21182 \dots

Solution steps

\int_{5}^{\infty} \frac{1}{x ^{2} - 4} d x

Compute the indefinite integral:

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

Compute the boundaries:

\int_{5}^{\infty} \frac{1}{x ^{2} - 4} d x = 0 - (- \frac{ln ( \frac{7}{2} ) - ln ( \frac{3}{2} )}{4})

= 0 - (- \frac{ln ( \frac{7}{2} ) - ln ( \frac{3}{2} )}{4})

Simplify

= \frac{ln ( \frac{7}{2} ) - ln ( \frac{3}{2} )}{4}

\frac{d}{d x} (\frac{1}{4 x ^{2} + 2 x})

x \to 2 lim (2 + 3)

\frac{d}{d x} (\frac{x ^{3} + 3 x ^{2} - 5 x + 2}{3})

y^{^{'''}} - 2 y^{^{''}} - 3 y^{^{'}} = 0

x \to - 1 + lim (\frac{- 1}{x ^{2} - 1})

What is the integral from 5 to infinity of 1/(x^2-4) ?
The integral from 5 to infinity of 1/(x^2-4) is (ln(7/2)-ln(3/2))/4