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

\int_{0}^{1} (5 x^{4} - 4 x^{3}) d x

g (x) = 2 x^{2}

h \to 0 lim (\frac{cos ( h )}{h})

\int \frac{3 x}{9 + x ^{2}} d x

f (x) = 2^{2 x}

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