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

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

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

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

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

Solution steps

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

Take the constant out:

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

= 3 \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 )}

= 3 \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

= 3 (\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))

= 3 (\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}

= 3 \cdot \frac{- ln ( 2 ) + ln ( 5 )}{8}

\frac{\partial}{\partial x} (x^{\frac{1}{3}})

n = 1 \sum \infty (- 1)^{n + 1} n

\frac{d}{d x} (x + 1 + 1)

f (x) = \frac{5 x}{x ^{2} + 1}, at x = 0

\frac{d y}{d θ} = sin^{4} (π θ)

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