What is the integral from-4 to 2 of (2x)/(3x^2+10) ?

The integral from-4 to 2 of (2x)/(3x^2+10) is -1/3 ln(29/11)

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integral from-4 to 2 of (2x)/(3x^2+10)

\int_{- 4}^{2} \frac{2 x}{3 x ^{2} + 10} d x

- \frac{1}{3} ln (\frac{29}{11})

Decimal

- 0.32313 \dots

Solution steps

\int_{- 4}^{2} \frac{2 x}{3 x ^{2} + 10} d x

Take the constant out:

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

= 2 \cdot \int_{- 4}^{2} \frac{x}{3 x ^{2} + 10} d x

Apply u-substitution

: \int_{58}^{22} \frac{1}{6 u} d u

= 2 \cdot \int_{58}^{22} \frac{1}{6 u} d u

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

= 2 (- \int_{22}^{58} \frac{1}{6 u} d u)

Take the constant out:

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

= 2 (- \frac{1}{6} \cdot \int_{22}^{58} \frac{1}{u} d u)

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= 2 (- \frac{1}{6} [ln ∣ u ∣]_{22}^{58})

Simplify

2 (- \frac{1}{6} [ln ∣ u ∣]_{22}^{58}) : - \frac{1}{3} [ln ∣ u ∣]_{22}^{58}

= - \frac{1}{3} [ln ∣ u ∣]_{22}^{58}

Compute the boundaries:

ln (58) - ln (22)

= - \frac{1}{3} (ln (58) - ln (22))

Simplify

= - \frac{1}{3} ln (\frac{29}{11})

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What is the integral from-4 to 2 of (2x)/(3x^2+10) ?
The integral from-4 to 2 of (2x)/(3x^2+10) is -1/3 ln(29/11)