What is the integral from 1 to 2 of (4/x-(16)/(x^2)) ?

The integral from 1 to 2 of (4/x-(16)/(x^2)) is 4ln(2)-8

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integral from 1 to 2 of (4/x-(16)/(x^2))

\int_{1}^{2} (\frac{4}{x} - \frac{16}{x ^{2}}) d x

4 ln (2) - 8

Decimal

- 5.22741 \dots

Solution steps

\int_{1}^{2} \frac{4}{x} - \frac{16}{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{4}{x} d x - \int_{1}^{2} \frac{16}{x ^{2}} d x

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

\int_{1}^{2} \frac{16}{x ^{2}} d x = 8

= 4 ln (2) - 8

What is the integral from 1 to 2 of (4/x-(16)/(x^2)) ?
The integral from 1 to 2 of (4/x-(16)/(x^2)) is 4ln(2)-8