What is the integral of (2x^4-8x^3-6x^2+4)/(x^3) ?

The integral of (2x^4-8x^3-6x^2+4)/(x^3) is x^2-8x-6ln|x|-2/(x^2)+C

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integral of (2x^4-8x^3-6x^2+4)/(x^3)

\int \frac{2 x ^{4} - 8 x ^{3} - 6 x ^{2} + 4}{x ^{3}} d x

x^{2} - 8 x - 6 ln ∣ x ∣ - \frac{2}{x ^{2}} + C

Solution steps

\int \frac{2 x ^{4} - 8 x ^{3} - 6 x ^{2} + 4}{x ^{3}} d x

Expand

\frac{2 x ^{4} - 8 x ^{3} - 6 x ^{2} + 4}{x ^{3}} : 2 x - 8 - \frac{6}{x} + \frac{4}{x ^{3}}

= \int 2 x - 8 - \frac{6}{x} + \frac{4}{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 2 x d x - \int 8 d x - \int \frac{6}{x} d x + \int \frac{4}{x ^{3}} d x

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

\int 8 d x = 8 x

\int \frac{6}{x} d x = 6 ln ∣ x ∣

\int \frac{4}{x ^{3}} d x = - \frac{2}{x ^{2}}

= x^{2} - 8 x - 6 ln ∣ x ∣ - \frac{2}{x ^{2}}

Add a constant to the solution

= x^{2} - 8 x - 6 ln ∣ x ∣ - \frac{2}{x ^{2}} + C

y^{^{''}} - 4 y^{^{'}} + 4 y = t^{- 5} e^{2 t}

x \to - 8 lim (x + 5)

\int \frac{1}{cos ^{2} ( 6 x )} d x

\frac{\partial}{\partial y} ((x^{2} + y) (x - y^{2}))

\int_{0}^{1} π (3 x)^{2} d x

What is the integral of (2x^4-8x^3-6x^2+4)/(x^3) ?
The integral of (2x^4-8x^3-6x^2+4)/(x^3) is x^2-8x-6ln|x|-2/(x^2)+C