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

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

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

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

- ln (5) + 3 ln (2)

Decimal Notation

0.47000 \dots

Solution steps

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

Take the constant out:

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

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

Take the partial fraction of

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

= 2 \cdot \int_{1}^{2} \frac{1}{x} - \frac{x}{x ^{2} + 1} 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

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

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

\int_{1}^{2} \frac{x}{x ^{2} + 1} d x = \frac{1}{2} (ln (5) - ln (2))

= 2 (ln (2) - \frac{1}{2} (ln (5) - ln (2)))

Simplify

2 (ln (2) - \frac{1}{2} (ln (5) - ln (2))) : - ln (5) + 3 ln (2)

= - ln (5) + 3 ln (2)

e^{3 t}

x \to 2 lim (x 4 - x^{2})

\int x^{2} + \frac{1}{( 3 x ) ^{2}} d x

\frac{d}{d x} (2 x^{3} - 15 x^{2} + 36 x + 5)

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