What is the integral of ((5-x)/x)^2 ?

The integral of ((5-x)/x)^2 is -(25)/x-10ln|x|+x+C

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integral of ((5-x)/x)^2

\int (\frac{5 - x}{x})^{2} d x

- \frac{25}{x} - 10 ln ∣ x ∣ + x + C

Solution steps

\int (\frac{5 - x}{x})^{2} d x

Apply exponent rule:

(\frac{a}{b})^{c} = \frac{a ^{c}}{b ^{c}}

= \int \frac{( 5 - x ) ^{2}}{x ^{2}} d x

Expand

\frac{( 5 - x ) ^{2}}{x ^{2}} : \frac{25}{x ^{2}} - \frac{10}{x} + 1

= \int \frac{25}{x ^{2}} - \frac{10}{x} + 1 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 \frac{25}{x ^{2}} d x - \int \frac{10}{x} d x + \int 1 d x

\int \frac{25}{x ^{2}} d x = - \frac{25}{x}

\int \frac{10}{x} d x = 10 ln ∣ x ∣

\int 1 d x = x

= - \frac{25}{x} - 10 ln ∣ x ∣ + x

Add a constant to the solution

= - \frac{25}{x} - 10 ln ∣ x ∣ + x + C

\int x (1 - x) d x

\frac{d}{d x} (ln (ln (14 x)))

\frac{\partial}{\partial x} (∣ x y ∣)

\frac{d y}{d x} + \frac{y}{x} = x^{2}

t^{2} + 9

What is the integral of ((5-x)/x)^2 ?
The integral of ((5-x)/x)^2 is -(25)/x-10ln|x|+x+C