What is the integral from 0 to 2 of 1/(x^2-4) ?

The integral from 0 to 2 of 1/(x^2-4) is diverges

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integral from 0 to 2 of 1/(x^2-4)

\int_{0}^{2} \frac{1}{x ^{2} - 4} d x

diverges

Solution steps

\int_{0}^{2} \frac{1}{x ^{2} - 4} d x

Apply Integral Substitution

= \int_{0}^{1} \frac{1}{2 ( u ^{2} - 1 )} d u

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= \frac{1}{2} \cdot \int_{0}^{1} \frac{1}{u ^{2} - 1} d u

Factor

u^{2} - 1 : - (- u^{2} + 1)

= \frac{1}{2} \cdot \int_{0}^{1} \frac{1}{- ( - u ^{2} + 1 )} d u

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= \frac{1}{2} (- \int_{0}^{1} \frac{1}{- u ^{2} + 1} d u)

Use the common integral:

\int \frac{1}{- u ^{2} + 1} d u = \frac{ln ∣ u + 1 ∣}{2} - \frac{ln ∣ u - 1 ∣}{2}

= \frac{1}{2} (- [\frac{ln ∣ u + 1 ∣}{2} - \frac{ln ∣ u - 1 ∣}{2}]_{0}^{1})

Simplify

\frac{1}{2} (- [\frac{ln ∣ u + 1 ∣}{2} - \frac{ln ∣ u - 1 ∣}{2}]_{0}^{1}) : - \frac{1}{2} [\frac{ln ∣ u + 1 ∣ - ln ∣ u - 1 ∣}{2}]_{0}^{1}

= - \frac{1}{2} [\frac{ln ∣ u + 1 ∣ - ln ∣ u - 1 ∣}{2}]_{0}^{1}

Compute the boundaries:

diverges

= diverges

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

\int θ^{2} sin (2 θ) d θ

\int e^{- s t} t^{3} d t

\frac{d}{d x} ((sin (x)) d x)

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

What is the integral from 0 to 2 of 1/(x^2-4) ?
The integral from 0 to 2 of 1/(x^2-4) is diverges