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

The integral from 2 to 5 of (2x^3-4x+1/x) is ln(5)-ln(2)+525/2

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

\int_{2}^{5} (2 x^{3} - 4 x + \frac{1}{x}) d x

ln (5) - ln (2) + \frac{525}{2}

Decimal Notation

263.41629 \dots

Solution steps

\int_{2}^{5} 2 x^{3} - 4 x + \frac{1}{x} 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

= \int_{2}^{5} 2 x^{3} d x - \int_{2}^{5} 4 x d x + \int_{2}^{5} \frac{1}{x} d x

\int_{2}^{5} 2 x^{3} d x = \frac{609}{2}

\int_{2}^{5} 4 x d x = 42

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

= \frac{609}{2} - 42 + ln (5) - ln (2)

Combine the fractions

- 42 + \frac{609}{2} : \frac{525}{2}

= ln (5) - ln (2) + \frac{525}{2}

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\int_{1}^{2} cos (x - 1) d x

What is the integral from 2 to 5 of (2x^3-4x+1/x) ?
The integral from 2 to 5 of (2x^3-4x+1/x) is ln(5)-ln(2)+525/2