What is the integral from 1 to 3 of x/(3x^2+1) ?

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

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integral from 1 to 3 of x/(3x^2+1)

\int_{1}^{3} \frac{x}{3 x ^{2} + 1} d x

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

Decimal Notation

0.32431 \dots

Solution steps

\int_{1}^{3} \frac{x}{3 x ^{2} + 1} d x

Apply u-substitution

: \int_{4}^{28} \frac{1}{6 u} d u

= \int_{4}^{28} \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_{4}^{28} \frac{1}{u} d u

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= \frac{1}{6} [ln ∣ u ∣]_{4}^{28}

Compute the boundaries:

ln (28) - 2 ln (2)

= \frac{1}{6} (ln (28) - 2 ln (2))

Simplify

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

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