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

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

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integral from 2 to e of (x^4-5x)/(x^2)

\int_{2}^{e} \frac{x ^{4} - 5 x}{x ^{2}} d x

\frac{e ^{3} - 8}{3} - 5 (1 - ln (2))

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2.49424 \dots

Solution steps

\int_{2}^{e} \frac{x ^{4} - 5 x}{x ^{2}} d x

Expand

\frac{x ^{4} - 5 x}{x ^{2}} : x^{2} - \frac{5}{x}

= \int_{2}^{e} x^{2} - \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_{2}^{e} x^{2} d x - \int_{2}^{e} \frac{5}{x} d x

\int_{2}^{e} x^{2} d x = \frac{e ^{3} - 8}{3}

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

= \frac{e ^{3} - 8}{3} - 5 (1 - ln (2))

\int_{0}^{1} x^{- \frac{1}{2}} ln (x) d x

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

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