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

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

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

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

ln (2) - 2

Decimal Notation

- 1.30685 \dots

Solution steps

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

Expand

\frac{x - 4}{x ^{2}} : \frac{1}{x} - \frac{4}{x ^{2}}

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

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

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

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

= ln (2) - 2

f (x) = 2 x^{3} - 2 x, (1, 0)

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