What is the integral from 1 to 6 of (x^4-3x^2)/(x^3) ?

The integral from 1 to 6 of (x^4-3x^2)/(x^3) is 35/2-3ln(6)

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integral from 1 to 6 of (x^4-3x^2)/(x^3)

\int_{1}^{6} \frac{x ^{4} - 3 x ^{2}}{x ^{3}} d x

\frac{35}{2} - 3 ln (6)

Decimal

12.12472 \dots

Solution steps

\int_{1}^{6} \frac{x ^{4} - 3 x ^{2}}{x ^{3}} d x

Simplify

\frac{x ^{4} - 3 x ^{2}}{x ^{3}} : \frac{x ^{2} - 3}{x}

= \int_{1}^{6} \frac{x ^{2} - 3}{x} d x

Expand

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

= \int_{1}^{6} x - \frac{3}{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}^{6} x d x - \int_{1}^{6} \frac{3}{x} d x

\int_{1}^{6} x d x = \frac{35}{2}

\int_{1}^{6} \frac{3}{x} d x = 3 ln (6)

= \frac{35}{2} - 3 ln (6)

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What is the integral from 1 to 6 of (x^4-3x^2)/(x^3) ?
The integral from 1 to 6 of (x^4-3x^2)/(x^3) is 35/2-3ln(6)