What is the integral from 5 to 10 of (6x)/(3x^2-5) ?

The integral from 5 to 10 of (6x)/(3x^2-5) is ln(59/14)

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integral from 5 to 10 of (6x)/(3x^2-5)

\int_{5}^{10} \frac{6 x}{3 x ^{2} - 5} d x

ln (\frac{59}{14})

Solution steps

\int_{5}^{10} \frac{6 x}{3 x ^{2} - 5} d x

Take the constant out:

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

= 6 \cdot \int_{5}^{10} \frac{x}{3 x ^{2} - 5} d x

Apply u-substitution

: \int_{70}^{295} \frac{1}{6 u} d u

= 6 \cdot \int_{70}^{295} \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_{70}^{295} \frac{1}{u} d u

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= 6 \cdot \frac{1}{6} [ln ∣ u ∣]_{70}^{295}

Simplify

6 \cdot \frac{1}{6} [ln ∣ u ∣]_{70}^{295} : [ln ∣ u ∣]_{70}^{295}

= [ln ∣ u ∣]_{70}^{295}

Compute the boundaries:

ln (295) - ln (70)

= ln (295) - ln (70)

Simplify

= ln (\frac{59}{14})

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What is the integral from 5 to 10 of (6x)/(3x^2-5) ?
The integral from 5 to 10 of (6x)/(3x^2-5) is ln(59/14)