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

The integral from 0 to 5 of x/(3x^2+3) is 1/6 ln(26)

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

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

\frac{1}{6} ln (26)

Decimal

0.54301 \dots

Solution steps

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

Apply u-substitution

: \int_{3}^{78} \frac{1}{6 u} d u

= \int_{3}^{78} \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_{3}^{78} \frac{1}{u} d u

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= \frac{1}{6} [ln ∣ u ∣]_{3}^{78}

Compute the boundaries:

ln (78) - ln (3)

= \frac{1}{6} (ln (78) - ln (3))

Simplify

= \frac{1}{6} ln (26)

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What is the integral from 0 to 5 of x/(3x^2+3) ?
The integral from 0 to 5 of x/(3x^2+3) is 1/6 ln(26)