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

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

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

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

diverges

Solution steps

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

Take the constant out:

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

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

Apply u-substitution

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

= 2 \cdot \int_{0}^{- 4} \frac{1}{2 u} d u

\int_{a}^{b} f (x) d x = - \int_{b}^{a} f (x) d x, a < b

= 2 (- \int_{- 4}^{0} \frac{1}{2 u} d u)

Take the constant out:

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

= 2 (- \frac{1}{2} \cdot \int_{- 4}^{0} \frac{1}{u} d u)

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= 2 (- \frac{1}{2} [ln ∣ u ∣]_{- 4}^{0})

Simplify

2 (- \frac{1}{2} [ln ∣ u ∣]_{- 4}^{0}) : - [ln ∣ u ∣]_{- 4}^{0}

= - [ln ∣ u ∣]_{- 4}^{0}

Compute the boundaries:

diverges

= diverges

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