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

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

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

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

x^{4} - 3 x^{2} + C

Solution steps

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

Expand

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

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

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

= x^{4} - 3 x^{2}

Add a constant to the solution

= x^{4} - 3 x^{2} + C

\int (x^{2} ln (2 x)) d x

e^{x^{2}} - 1

\int (\frac{x}{2})^{4} d x

y^{^{'}} = \frac{y}{x}

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