What is the integral from 5 to 2 of ((12-x^3))/(x^4) ?

The integral from 5 to 2 of ((12-x^3))/(x^4) is -117/250+ln(5)-ln(2)

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integral from 5 to 2 of ((12-x^3))/(x^4)

\int_{5}^{2} \frac{( 12 - x ^{3} )}{x ^{4}} d x

- \frac{117}{250} + ln (5) - ln (2)

Decimal Notation

0.44829 \dots

Solution steps

\int_{5}^{2} \frac{12 - x ^{3}}{x ^{4}} d x

\int_{a}^{b} f (x) d x = - \int_{b}^{a} f (x) d x, a < b

= - \int_{2}^{5} \frac{12 - x ^{3}}{x ^{4}} d x

Expand

\frac{12 - x ^{3}}{x ^{4}} : \frac{12}{x ^{4}} - \frac{1}{x}

= - \int_{2}^{5} \frac{12}{x ^{4}} - \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} \frac{12}{x ^{4}} d x - \int_{2}^{5} \frac{1}{x} d x)

\int_{2}^{5} \frac{12}{x ^{4}} d x = \frac{117}{250}

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

= - (\frac{117}{250} - (ln (5) - ln (2)))

Simplify

= - \frac{117}{250} + ln (5) - ln (2)

\int_{1}^{3} e^{3 x} d x

\int_{0}^{2} 3 x^{2} + 2 x + 1 d x

\int_{1}^{x^{2}} \frac{1}{t ^{5}} d t

\int_{- 1}^{5} 9 y + 9 y^{3} + \frac{3 y ^{3}}{4} d y

\int_{- \infty}^{- 1} t^{3} d t

What is the integral from 5 to 2 of ((12-x^3))/(x^4) ?
The integral from 5 to 2 of ((12-x^3))/(x^4) is -117/250+ln(5)-ln(2)