What is the integral from 1 to 2 of 7/(x^3+4x) ?

The integral from 1 to 2 of 7/(x^3+4x) is 7*(-ln(2)+ln(5))/8

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integral from 1 to 2 of 7/(x^3+4x)

\int_{1}^{2} \frac{7}{x ^{3} + 4 x} d x

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

Decimal Notation

0.80175 \dots

Solution steps

\int_{1}^{2} \frac{7}{x ^{3} + 4 x} d x

Take the constant out:

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

= 7 \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 )}

= 7 \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

= 7 (\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))

= 7 (\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}

= 7 \cdot \frac{- ln ( 2 ) + ln ( 5 )}{8}

\frac{\partial}{\partial y} (sin (y) - y sin (x))

\frac{d}{d x} (e^{x x^{2} + 1})

t \to 1 - lim (\frac{2 t}{t ^{2} - 1})

y^{^{''}} - 2 y + 4 y = 0

\int 11 x^{10} - \frac{10}{x ^{10}} d x

What is the integral from 1 to 2 of 7/(x^3+4x) ?
The integral from 1 to 2 of 7/(x^3+4x) is 7*(-ln(2)+ln(5))/8