What is the integral from 5 to 9 of ((x-4)/(x^2)) ?

The integral from 5 to 9 of ((x-4)/(x^2)) is 2ln(3)-ln(5)-16/45

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integral from 5 to 9 of ((x-4)/(x^2))

\int_{5}^{9} (\frac{x - 4}{x ^{2}}) d x

2 ln (3) - ln (5) - \frac{16}{45}

Decimal Notation

0.23223 \dots

Solution steps

\int_{5}^{9} \frac{x - 4}{x ^{2}} d x

Expand

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

= \int_{5}^{9} \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_{5}^{9} \frac{1}{x} d x - \int_{5}^{9} \frac{4}{x ^{2}} d x

\int_{5}^{9} \frac{1}{x} d x = 2 ln (3) - ln (5)

\int_{5}^{9} \frac{4}{x ^{2}} d x = \frac{16}{45}

= 2 ln (3) - ln (5) - \frac{16}{45}

\frac{d}{d x} (\frac{x ^{2} + 4 x + 5 x ^{3}}{x + 1})

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

\int_{0}^{1} \frac{25}{4 y - 1} d y

x \to 0 + lim (x ln (x + x^{2}))

\frac{d P}{d x} = P - P^{2}

What is the integral from 5 to 9 of ((x-4)/(x^2)) ?
The integral from 5 to 9 of ((x-4)/(x^2)) is 2ln(3)-ln(5)-16/45