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

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

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

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

\frac{- ln ( 2 ) + ln ( 5 )}{4}

Decimal Notation

0.22907 \dots

Solution steps

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

= 2 \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

= 2 (\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))

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

Simplify

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

= \frac{- ln ( 2 ) + ln ( 5 )}{4}

\frac{\partial}{\partial y} (x^{2} \cdot y^{2})

y^{^{''}} - 7 y^{^{'}} - 18 y = 0

x \to 2 lim (\frac{( x + 2 )}{3 x})

\frac{\partial}{\partial y} (y^{2} cos (x))

\frac{\partial}{\partial x} (6 tan (x y))

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