What is the integral from 1 to 2 of (11)/(x^3+4x) ?

The integral from 1 to 2 of (11)/(x^3+4x) is 11*(-ln(2)+ln(5))/8

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integral from 1 to 2 of (11)/(x^3+4x)

\int_{1}^{2} \frac{11}{x ^{3} + 4 x} d x

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

Decimal Notation

1.25989 \dots

Solution steps

\int_{1}^{2} \frac{11}{x ^{3} + 4 x} d x

Take the constant out:

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

= 11 \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 )}

= 11 \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

= 11 (\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))

= 11 (\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}

= 11 \cdot \frac{- ln ( 2 ) + ln ( 5 )}{8}

2 cos^{2} (t)

\int x^{4} e^{x^{5} + 2} d x

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

\int_{1}^{4} x ln (x) d x

\int \frac{x}{5 - 4 x ^{2} - x ^{4}} d x

What is the integral from 1 to 2 of (11)/(x^3+4x) ?
The integral from 1 to 2 of (11)/(x^3+4x) is 11*(-ln(2)+ln(5))/8