What is the integral from 1 to 2 of 9/(x^3+2x) ?

The integral from 1 to 2 of 9/(x^3+2x) is 9/4 ln(2)

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integral from 1 to 2 of 9/(x^3+2x)

\int_{1}^{2} \frac{9}{x ^{3} + 2 x} d x

\frac{9}{4} ln (2)

Solution steps

\int_{1}^{2} \frac{9}{x ^{3} + 2 x} d x

Take the constant out:

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

= 9 \cdot \int_{1}^{2} \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 )}

= 9 \cdot \int_{1}^{2} \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

= 9 (\int_{1}^{2} \frac{1}{2 x} d x - \int_{1}^{2} \frac{x}{2 ( x ^{2} + 2 )} d x)

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

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

= 9 (\frac{1}{2} ln (2) - \frac{1}{4} ln (2))

Simplify

9 (\frac{1}{2} ln (2) - \frac{1}{4} ln (2)) : \frac{9}{4} ln (2)

= \frac{9}{4} ln (2)

y^{^{''}} + 5 y^{^{'}} + 6 y = 0

y^{^{''}} + 3 y = 9

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

\int 4 r^{2} + 1 r d r

\frac{d}{d x} (arccos (\frac{x}{8}))

What is the integral from 1 to 2 of 9/(x^3+2x) ?
The integral from 1 to 2 of 9/(x^3+2x) is 9/4 ln(2)