What is the integral of (9+x)/(x^2) ?

The integral of (9+x)/(x^2) is -9/x+ln|x|+C

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integral of (9+x)/(x^2)

\int \frac{9 + x}{x ^{2}} d x

- \frac{9}{x} + ln ∣ x ∣ + C

Solution steps

\int \frac{9 + x}{x ^{2}} d x

Expand

\frac{9 + x}{x ^{2}} : \frac{9}{x ^{2}} + \frac{1}{x}

= \int \frac{9}{x ^{2}} + \frac{1}{x} 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{9}{x ^{2}} d x + \int \frac{1}{x} d x

\int \frac{9}{x ^{2}} d x = - \frac{9}{x}

\int \frac{1}{x} d x = ln ∣ x ∣

= - \frac{9}{x} + ln ∣ x ∣

Add a constant to the solution

= - \frac{9}{x} + ln ∣ x ∣ + C

f (x) = - 3 - 7 x^{2}, at x = 42

\frac{d}{d x} (e^{- 0.05 x})

f (x) = \frac{1}{x} + 4, at x = - 4

\frac{d y}{d x} + \frac{- x}{x + 1} = 0

\frac{d y}{d x} = - \frac{x y}{y - 1}

What is the integral of (9+x)/(x^2) ?
The integral of (9+x)/(x^2) is -9/x+ln|x|+C