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

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

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

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

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

Decimal

0.45814 \dots

Solution steps

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

Take the constant out:

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

= 4 \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 )}

= 4 \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

= 4 (\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))

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

Simplify

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

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

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

\int csc (\frac{a x}{b}) \cdot cot (\frac{a x}{b}) d x

f (x) = 4 x - 3

\int_{0}^{2} e^{- 2 x} + 4 d x

\frac{d}{d x} (\frac{6}{2 x + 1})

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