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

The integral of ((4x^8-6x^4))/((2x^6)) is (2x^4+9)/(3x)+C

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

\int \frac{( 4 x ^{8} - 6 x ^{4} )}{( 2 x ^{6} )} d x

\frac{2 x ^{4} + 9}{3 x} + C

Solution steps

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

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= \frac{1}{2} \cdot \int \frac{4 x ^{8} - 6 x ^{4}}{x ^{6}} d x

Expand

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

= \frac{1}{2} \cdot \int 4 x^{2} - \frac{6}{x ^{2}} 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

= \frac{1}{2} (\int 4 x^{2} d x - \int \frac{6}{x ^{2}} d x)

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

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

= \frac{1}{2} (\frac{4 x ^{3}}{3} - (- \frac{6}{x}))

Simplify

\frac{1}{2} (\frac{4 x ^{3}}{3} - (- \frac{6}{x})) : \frac{2 x ^{4} + 9}{3 x}

= \frac{2 x ^{4} + 9}{3 x}

Add a constant to the solution

= \frac{2 x ^{4} + 9}{3 x} + C

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What is the integral of ((4x^8-6x^4))/((2x^6)) ?
The integral of ((4x^8-6x^4))/((2x^6)) is (2x^4+9)/(3x)+C