What is the integral from 1 to 4 of (x^3-6x+5)/x ?

The integral from 1 to 4 of (x^3-6x+5)/x is 10ln(2)+3

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integral from 1 to 4 of (x^3-6x+5)/x

\int_{1}^{4} \frac{x ^{3} - 6 x + 5}{x} d x

10 ln (2) + 3

Decimal Notation

9.93147 \dots

Solution steps

\int_{1}^{4} \frac{x ^{3} - 6 x + 5}{x} d x

Expand

\frac{x ^{3} - 6 x + 5}{x} : x^{2} - 6 + \frac{5}{x}

= \int_{1}^{4} x^{2} - 6 + \frac{5}{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_{1}^{4} x^{2} d x - \int_{1}^{4} 6 d x + \int_{1}^{4} \frac{5}{x} d x

\int_{1}^{4} x^{2} d x = 21

\int_{1}^{4} 6 d x = 18

\int_{1}^{4} \frac{5}{x} d x = 10 ln (2)

= 21 - 18 + 10 ln (2)

Simplify

= 10 ln (2) + 3

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\int_{1.5}^{2} 2 - x d x

What is the integral from 1 to 4 of (x^3-6x+5)/x ?
The integral from 1 to 4 of (x^3-6x+5)/x is 10ln(2)+3