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

The integral from 1 to 4 of (x^5-x)/(3x^3) is 27/4

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

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

\frac{27}{4}

Decimal Notation

6.75

Solution steps

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

Take the constant out:

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

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

Expand

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

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

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

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

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

= \frac{1}{3} (21 - \frac{3}{4})

Simplify

\frac{1}{3} (21 - \frac{3}{4}) : \frac{27}{4}

= \frac{27}{4}

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What is the integral from 1 to 4 of (x^5-x)/(3x^3) ?
The integral from 1 to 4 of (x^5-x)/(3x^3) is 27/4