What is the integral of ((x^5-x^3+6x))/(x^4) ?

The integral of ((x^5-x^3+6x))/(x^4) is (x^2)/2-ln|x|-3/(x^2)+C

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integral of ((x^5-x^3+6x))/(x^4)

\int \frac{( x ^{5} - x ^{3} + 6 x )}{x ^{4}} d x

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

Solution steps

\int \frac{x ^{5} - x ^{3} + 6 x}{x ^{4}} d x

Expand

\frac{x ^{5} - x ^{3} + 6 x}{x ^{4}} : x - \frac{1}{x} + \frac{6}{x ^{3}}

= \int x - \frac{1}{x} + \frac{6}{x ^{3}} 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{1}{x} d x + \int \frac{6}{x ^{3}} d x

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

\int \frac{1}{x} d x = ln ∣ x ∣

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

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

Add a constant to the solution

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

x^{3} - 11 x^{2} + 30 x, - x^{3} + 11 x^{2} - 30 x

f (x) = (x - 2)^{3} (x - 1)

y^{^{''}} + 36 y = - 24 sin (6 t)

(x e^{x})^{^{'}}

(e^{x} + y) d x + (e^{y} + x) d y = 0

What is the integral of ((x^5-x^3+6x))/(x^4) ?
The integral of ((x^5-x^3+6x))/(x^4) is (x^2)/2-ln|x|-3/(x^2)+C