What is the integral from 0 to 5 of x/(3x^2+6) ?

The integral from 0 to 5 of x/(3x^2+6) is 1/6 ln(27/2)

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integral from 0 to 5 of x/(3x^2+6)

\int_{0}^{5} \frac{x}{3 x ^{2} + 6} d x

\frac{1}{6} ln (\frac{27}{2})

Solution steps

\int_{0}^{5} \frac{x}{3 x ^{2} + 6} d x

Apply u-substitution

: \int_{6}^{81} \frac{1}{6 u} d u

= \int_{6}^{81} \frac{1}{6 u} d u

Take the constant out:

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

= \frac{1}{6} \cdot \int_{6}^{81} \frac{1}{u} d u

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= \frac{1}{6} [ln ∣ u ∣]_{6}^{81}

Compute the boundaries:

4 ln (3) - ln (6)

= \frac{1}{6} (4 ln (3) - ln (6))

Simplify

= \frac{1}{6} ln (\frac{27}{2})

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What is the integral from 0 to 5 of x/(3x^2+6) ?
The integral from 0 to 5 of x/(3x^2+6) is 1/6 ln(27/2)