What is the integral from 1 to 2 of 7/(x^3+5x) ?

The integral from 1 to 2 of 7/(x^3+5x) is 7(1/5 ln(2)-1/10 ln(3/2))

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integral from 1 to 2 of 7/(x^3+5x)

\int_{1}^{2} \frac{7}{x ^{3} + 5 x} d x

7 (\frac{1}{5} ln (2) - \frac{1}{10} ln (\frac{3}{2}))

Decimal

0.68658 \dots

Solution steps

\int_{1}^{2} \frac{7}{x ^{3} + 5 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} + 5 x} d x

Take the partial fraction of

\frac{1}{x ^{3} + 5 x} : \frac{1}{5 x} - \frac{x}{5 ( x ^{2} + 5 )}

= 7 \cdot \int_{1}^{2} \frac{1}{5 x} - \frac{x}{5 ( x ^{2} + 5 )} 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}{5 x} d x - \int_{1}^{2} \frac{x}{5 ( x ^{2} + 5 )} d x)

\int_{1}^{2} \frac{1}{5 x} d x = \frac{1}{5} ln (2)

\int_{1}^{2} \frac{x}{5 ( x ^{2} + 5 )} d x = \frac{1}{10} ln (\frac{3}{2})

= 7 (\frac{1}{5} ln (2) - \frac{1}{10} ln (\frac{3}{2}))

\int x cos^{2} (6 x) d x

\frac{d}{d x} (0.5 x^{2} e^{- 0.6 x})

x ln (x) y^{^{'}} + y = 8 x e^{x}

\frac{x}{1 - x ^{2}}

n = 0 \sum \infty \frac{1}{n ln ( n )}

What is the integral from 1 to 2 of 7/(x^3+5x) ?
The integral from 1 to 2 of 7/(x^3+5x) is 7(1/5 ln(2)-1/10 ln(3/2))