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

The integral from 2 to 3 of (x^2+1)/(x^3+3x) is 1/3 ln(18/7)

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

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

\frac{1}{3} ln (\frac{18}{7})

Decimal

0.31482 \dots

Solution steps

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

Apply u-substitution

= \int_{14}^{36} \frac{1}{3 u} d u

Take the constant out:

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

= \frac{1}{3} \cdot \int_{14}^{36} \frac{1}{u} d u

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= \frac{1}{3} [ln ∣ u ∣]_{14}^{36}

Compute the boundaries:

2 ln (6) - ln (14)

= \frac{1}{3} (2 ln (6) - ln (14))

Simplify

= \frac{1}{3} ln (\frac{18}{7})

What is the integral from 2 to 3 of (x^2+1)/(x^3+3x) ?
The integral from 2 to 3 of (x^2+1)/(x^3+3x) is 1/3 ln(18/7)