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

The integral from 1 to 2 of 9/(x^3+4x) is 9*(-ln(2)+ln(5))/8

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

\int_{1}^{2} \frac{9}{x ^{3} + 4 x} d x

9 \cdot \frac{- ln ( 2 ) + ln ( 5 )}{8}

Decimal Notation

1.03082 \dots

Solution steps

\int_{1}^{2} \frac{9}{x ^{3} + 4 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} + 4 x} d x

Take the partial fraction of

\frac{1}{x ^{3} + 4 x} : \frac{1}{4 x} - \frac{x}{4 ( x ^{2} + 4 )}

= 9 \cdot \int_{1}^{2} \frac{1}{4 x} - \frac{x}{4 ( x ^{2} + 4 )} 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}{4 x} d x - \int_{1}^{2} \frac{x}{4 ( x ^{2} + 4 )} d x)

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

\int_{1}^{2} \frac{x}{4 ( x ^{2} + 4 )} d x = \frac{1}{8} (3 ln (2) - ln (5))

= 9 (\frac{1}{4} ln (2) - \frac{1}{8} (3 ln (2) - ln (5)))

Join

\frac{1}{4} ln (2) - \frac{1}{8} (3 ln (2) - ln (5)) : \frac{- ln ( 2 ) + ln ( 5 )}{8}

= 9 \cdot \frac{- ln ( 2 ) + ln ( 5 )}{8}

\frac{d}{d x} (\frac{5}{2} x^{2})

y = - \frac{6}{x}, at x = - 4

\frac{d}{d x} (cot^{2} (3 x))

n = 5 \sum \infty e^{3 - 4 n}

y^{^{''}} + 4 π^{2} y = cos^{3} (2 π x)

What is the integral from 1 to 2 of 9/(x^3+4x) ?
The integral from 1 to 2 of 9/(x^3+4x) is 9*(-ln(2)+ln(5))/8