What is the integral from 1 to 4 of 1/(x(x-5)) ?

The integral from 1 to 4 of 1/(x(x-5)) is -4/5 ln(2)

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integral from 1 to 4 of 1/(x(x-5))

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

- \frac{4}{5} ln (2)

Decimal

- 0.55451 \dots

Solution steps

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

Take the partial fraction of

\frac{1}{x ( x - 5 )} : - \frac{1}{5 x} + \frac{1}{5 ( x - 5 )}

= \int_{1}^{4} - \frac{1}{5 x} + \frac{1}{5 ( x - 5 )} 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{1}{5 x} d x + \int_{1}^{4} \frac{1}{5 ( x - 5 )} d x

\int_{1}^{4} \frac{1}{5 x} d x = \frac{2}{5} ln (2)

\int_{1}^{4} \frac{1}{5 ( x - 5 )} d x = - \frac{2}{5} ln (2)

= - \frac{2}{5} ln (2) - \frac{2}{5} ln (2)

Simplify

- \frac{2}{5} ln (2) - \frac{2}{5} ln (2) : - \frac{4}{5} ln (2)

= - \frac{4}{5} ln (2)

y^{^{'}} (x^{2} + 1) + 3 x (y - 1) = 0

\int 7 x sin (2 x) d x

x \to 2 - lim (\frac{2}{2 - x})

t \to 2 lim (ln (t^{3}))

(\frac{1}{x}) d y = sin (x) (y^{3} + y) d x

What is the integral from 1 to 4 of 1/(x(x-5)) ?
The integral from 1 to 4 of 1/(x(x-5)) is -4/5 ln(2)