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

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

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

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

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

Solution steps

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

Expand

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

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

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

\int \frac{3}{x} d x = 3 ln ∣ x ∣

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

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

Simplify

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

Add a constant to the solution

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

x \to 3 - lim (\frac{x - 4}{x - 3})

n = 1 \sum \infty 2 - 3 (0.8)^{n}

\int_{0}^{\infty} x e^{- 2 x} d x

(x)^{4} cos (7 x - 5)

\int e^{- 5 x} sin (6 x) d x

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