What is the integral of (3x^2-6x^2+sqrt(x))/(x^3) ?

The integral of (3x^2-6x^2+sqrt(x))/(x^3) is -3ln|x|-2/(3x^{3/2)}+C

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integral of (3x^2-6x^2+sqrt(x))/(x^3)

\int \frac{3 x ^{2} - 6 x ^{2} + x}{x ^{3}} d x

- 3 ln ∣ x ∣ - \frac{2}{3 x ^{\frac{3}{2}}} + C

Solution steps

\int \frac{3 x ^{2} - 6 x ^{2} + x}{x ^{3}} d x

Simplify

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

Expand

\frac{- 3 x ^{2} + x}{x ^{3}} : - \frac{3}{x} + \frac{1}{x ^{\frac{5}{2}}}

= \int - \frac{3}{x} + \frac{1}{x ^{\frac{5}{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 \frac{3}{x} d x + \int \frac{1}{x ^{\frac{5}{2}}} d x

\int \frac{3}{x} d x = 3 ln ∣ x ∣

\int \frac{1}{x ^{\frac{5}{2}}} d x = - \frac{2}{3 x ^{\frac{3}{2}}}

= - 3 ln ∣ x ∣ - \frac{2}{3 x ^{\frac{3}{2}}}

Add a constant to the solution

= - 3 ln ∣ x ∣ - \frac{2}{3 x ^{\frac{3}{2}}} + C

y = 5 x + 20, at x = 1

f (x) = 16 x + 3

2 x^{2} y y^{^{'}} + 1 + 2 x y^{2} = 0

\int_{- 2}^{1} x ((4 - x^{2}) - (x + 2)) d x

What is the integral of (3x^2-6x^2+sqrt(x))/(x^3) ?
The integral of (3x^2-6x^2+sqrt(x))/(x^3) is -3ln|x|-2/(3x^{3/2)}+C