What is the integral of (sqrt(x)-3x)/(x^2) ?

The integral of (sqrt(x)-3x)/(x^2) is -2/(sqrt(x))-3ln|x|+C

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integral of (sqrt(x)-3x)/(x^2)

\int \frac{x - 3 x}{x ^{2}} d x

- \frac{2}{x} - 3 ln ∣ x ∣ + C

Solution steps

\int \frac{x - 3 x}{x ^{2}} d x

Expand

\frac{x - 3 x}{x ^{2}} : \frac{1}{x ^{\frac{3}{2}}} - \frac{3}{x}

= \int \frac{1}{x ^{\frac{3}{2}}} - \frac{3}{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{1}{x ^{\frac{3}{2}}} d x - \int \frac{3}{x} d x

\int \frac{1}{x ^{\frac{3}{2}}} d x = - \frac{2}{x}

\int \frac{3}{x} d x = 3 ln ∣ x ∣

= - \frac{2}{x} - 3 ln ∣ x ∣

Add a constant to the solution

= - \frac{2}{x} - 3 ln ∣ x ∣ + C

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What is the integral of (sqrt(x)-3x)/(x^2) ?
The integral of (sqrt(x)-3x)/(x^2) is -2/(sqrt(x))-3ln|x|+C