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

The integral from 1 to 2 of 7/(x^3+3x) is 7*(4ln(2)-ln(7))/6

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

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

7 \cdot \frac{4 ln ( 2 ) - ln ( 7 )}{6}

Decimal Notation

0.96445 \dots

Solution steps

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

Take the partial fraction of

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

= 7 \cdot \int_{1}^{2} \frac{1}{3 x} - \frac{x}{3 ( x ^{2} + 3 )} 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}{3 x} d x - \int_{1}^{2} \frac{x}{3 ( x ^{2} + 3 )} d x)

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

\int_{1}^{2} \frac{x}{3 ( x ^{2} + 3 )} d x = \frac{1}{6} (ln (7) - 2 ln (2))

= 7 (\frac{1}{3} ln (2) - \frac{1}{6} (ln (7) - 2 ln (2)))

Join

\frac{1}{3} ln (2) - \frac{1}{6} (ln (7) - 2 ln (2)) : \frac{4 ln ( 2 ) - ln ( 7 )}{6}

= 7 \cdot \frac{4 ln ( 2 ) - ln ( 7 )}{6}

y^{^{'''}} + 12 y^{^{''}} + 36 y^{^{'}} = 18 x^{2}

x^{3} + 3

f (x) = (3 + x^{5})^{\frac{2}{3}}

x^{3} cos (4 x)

\int \frac{tan ^{3} ( x )}{sec ^{6} ( x )} d x

What is the integral from 1 to 2 of 7/(x^3+3x) ?
The integral from 1 to 2 of 7/(x^3+3x) is 7*(4ln(2)-ln(7))/6