What is the integral from 3 to 9 of |1-9/x | ?

The integral from 3 to 9 of |1-9/x | is -6+9ln(3)

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integral from 3 to 9 of |1-9/x |

\int_{3}^{9} 1 - \frac{9}{x} d x

- 6 + 9 ln (3)

Decimal

3.88751 \dots

Solution steps

\int_{3}^{9} 1 - \frac{9}{x} d x

Eliminate Absolutes

= \int_{3}^{9} - 1 + \frac{9}{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_{3}^{9} 1 d x + \int_{3}^{9} \frac{9}{x} d x

\int_{3}^{9} 1 d x = 6

\int_{3}^{9} \frac{9}{x} d x = 9 ln (3)

= - 6 + 9 ln (3)

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

\int_{0}^{2} \frac{1}{x ^{2} - 2 x + 1} d x

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

\int_{0}^{3} 2 x (x - 5) d x

\int_{0}^{1} 3 y^{2} (- y + 1)^{2} d y

What is the integral from 3 to 9 of |1-9/x | ?
The integral from 3 to 9 of |1-9/x | is -6+9ln(3)