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

The integral from 2 to 3 of 1/(x^3+2x) is 1/2 (ln(3)-ln(2))-1/4 (ln(11)-ln(6))

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

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

\frac{1}{2} (ln (3) - ln (2)) - \frac{1}{4} (ln (11) - ln (6))

Decimal Notation

0.05119 \dots

Solution steps

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

Take the partial fraction of

\frac{1}{x ^{3} + 2 x} : \frac{1}{2 x} - \frac{x}{2 ( x ^{2} + 2 )}

= \int_{2}^{3} \frac{1}{2 x} - \frac{x}{2 ( x ^{2} + 2 )} d x

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int_{2}^{3} \frac{1}{2 x} d x - \int_{2}^{3} \frac{x}{2 ( x ^{2} + 2 )} d x

\int_{2}^{3} \frac{1}{2 x} d x = \frac{1}{2} (ln (3) - ln (2))

\int_{2}^{3} \frac{x}{2 ( x ^{2} + 2 )} d x = \frac{1}{4} (ln (11) - ln (6))

= \frac{1}{2} (ln (3) - ln (2)) - \frac{1}{4} (ln (11) - ln (6))

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

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\int_{0}^{\infty} x e^{- θ x} d x

What is the integral from 2 to 3 of 1/(x^3+2x) ?
The integral from 2 to 3 of 1/(x^3+2x) is 1/2 (ln(3)-ln(2))-1/4 (ln(11)-ln(6))