What is the integral from 2 to 3 of (2x)/(x^2-3) ?

The integral from 2 to 3 of (2x)/(x^2-3) is ln(6)

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integral from 2 to 3 of (2x)/(x^2-3)

\int_{2}^{3} \frac{2 x}{x ^{2} - 3} d x

ln (6)

Decimal

1.79175 \dots

Solution steps

\int_{2}^{3} \frac{2 x}{x ^{2} - 3} d x

Take the constant out:

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

= 2 \cdot \int_{2}^{3} \frac{x}{x ^{2} - 3} d x

Apply u-substitution

: \int_{1}^{6} \frac{1}{2 u} d u

= 2 \cdot \int_{1}^{6} \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 \cdot \frac{1}{2} \cdot \int_{1}^{6} \frac{1}{u} d u

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= 2 \cdot \frac{1}{2} [ln ∣ u ∣]_{1}^{6}

Simplify

2 \cdot \frac{1}{2} [ln ∣ u ∣]_{1}^{6} : [ln ∣ u ∣]_{1}^{6}

= [ln ∣ u ∣]_{1}^{6}

Compute the boundaries:

ln (6)

= ln (6)

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What is the integral from 2 to 3 of (2x)/(x^2-3) ?
The integral from 2 to 3 of (2x)/(x^2-3) is ln(6)