What is the integral of (x^3-8x)/(x^2) ?

The integral of (x^3-8x)/(x^2) is (x^2)/2-8ln|x|+C

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integral of (x^3-8x)/(x^2)

\int \frac{x ^{3} - 8 x}{x ^{2}} d x

\frac{x ^{2}}{2} - 8 ln ∣ x ∣ + C

Solution steps

\int \frac{x ^{3} - 8 x}{x ^{2}} d x

Expand

\frac{x ^{3} - 8 x}{x ^{2}} : x - \frac{8}{x}

= \int x - \frac{8}{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 x d x - \int \frac{8}{x} d x

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

\int \frac{8}{x} d x = 8 ln ∣ x ∣

= \frac{x ^{2}}{2} - 8 ln ∣ x ∣

Add a constant to the solution

= \frac{x ^{2}}{2} - 8 ln ∣ x ∣ + C

\frac{d}{d x} ((x^{2} + sin (x)) sec (x))

\frac{\partial}{\partial t} (s + t)

y^{^{''}} + 2 y^{^{'}} + 5 y = 3 sin (2 t)

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

\int \frac{x ^{4}}{x ^{2} + 3} d x

What is the integral of (x^3-8x)/(x^2) ?
The integral of (x^3-8x)/(x^2) is (x^2)/2-8ln|x|+C