What is the integral from 0 to 3 of x/(3x^2+7) ?

The integral from 0 to 3 of x/(3x^2+7) is 1/6 (ln(34)-ln(7))

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integral from 0 to 3 of x/(3x^2+7)

\int_{0}^{3} \frac{x}{3 x ^{2} + 7} d x

\frac{1}{6} (ln (34) - ln (7))

Solution steps

\int_{0}^{3} \frac{x}{3 x ^{2} + 7} d x

Apply u-substitution

: \int_{7}^{34} \frac{1}{6 u} d u

= \int_{7}^{34} \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_{7}^{34} \frac{1}{u} d u

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= \frac{1}{6} [ln ∣ u ∣]_{7}^{34}

Compute the boundaries:

ln (34) - ln (7)

= \frac{1}{6} (ln (34) - ln (7))

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What is the integral from 0 to 3 of x/(3x^2+7) ?
The integral from 0 to 3 of x/(3x^2+7) is 1/6 (ln(34)-ln(7))