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} (7 x + 5)

\int_{0}^{1} \frac{5 t ^{4}}{1 + t ^{5}} d t

\int \frac{1}{y ( y + 2 )} d y

- 2 sin (x)

\int \frac{1}{x ^{2} 1 + x ^{2}} d x

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