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

The integral from 1 to 4 of (4/x-(16)/(x^2)) is 8ln(2)-12

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

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

8 ln (2) - 12

Decimal Notation

- 6.45482 \dots

Solution steps

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

\int_{1}^{4} \frac{4}{x} d x = 8 ln (2)

\int_{1}^{4} \frac{16}{x ^{2}} d x = 12

= 8 ln (2) - 12

\frac{\partial}{\partial x} (\frac{3 x + 9}{2 - x})

\frac{d}{d x} (\frac{6 x}{x + 5})

f (x) = \frac{x ^{2}}{4}

x + x y^{^{'}} = y

What is the integral from 1 to 4 of (4/x-(16)/(x^2)) ?
The integral from 1 to 4 of (4/x-(16)/(x^2)) is 8ln(2)-12