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

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

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integral of (2x^4-2x^2)/(x^2)

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

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

Solution steps

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

Expand

\frac{2 x ^{4} - 2 x ^{2}}{x ^{2}} : \frac{2 x ^{4}}{x ^{2}} - \frac{2 x ^{2}}{x ^{2}}

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

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

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

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

Add a constant to the solution

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

ln (x + 1 + x^{2})

\frac{\partial}{\partial y} (x^{3} + x y)

x \to \infty lim (\frac{25 6 ^{\frac{x}{2}}}{4 ^{2 x}})

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

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