What is the integral from 1 to 2 of (6x)/(3x^2-2) ?

The integral from 1 to 2 of (6x)/(3x^2-2) is ln(10)

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integral from 1 to 2 of (6x)/(3x^2-2)

\int_{1}^{2} \frac{6 x}{3 x ^{2} - 2} d x

ln (10)

Decimal Notation

2.30258 \dots

Solution steps

\int_{1}^{2} \frac{6 x}{3 x ^{2} - 2} d x

Take the constant out:

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

= 6 \cdot \int_{1}^{2} \frac{x}{3 x ^{2} - 2} d x

Apply u-substitution

: \int_{1}^{10} \frac{1}{6 u} d u

= 6 \cdot \int_{1}^{10} \frac{1}{6 u} d u

Take the constant out:

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

= 6 \cdot \frac{1}{6} \cdot \int_{1}^{10} \frac{1}{u} d u

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= 6 \cdot \frac{1}{6} [ln ∣ u ∣]_{1}^{10}

Simplify

6 \cdot \frac{1}{6} [ln ∣ u ∣]_{1}^{10} : [ln ∣ u ∣]_{1}^{10}

= [ln ∣ u ∣]_{1}^{10}

Compute the boundaries:

ln (10)

= ln (10)

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What is the integral from 1 to 2 of (6x)/(3x^2-2) ?
The integral from 1 to 2 of (6x)/(3x^2-2) is ln(10)