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

The integral of (4x^3-6x^4)/(2x^6) is (3x-1)/(x^2)+C

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

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

\frac{3 x - 1}{x ^{2}} + C

Solution steps

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

Expand

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

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

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

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

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

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

Simplify

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

= \frac{3 x - 1}{x ^{2}}

Add a constant to the solution

= \frac{3 x - 1}{x ^{2}} + C

y^{^{'}} = 2 y - x^{2}

(\frac{x ^{2} + x + 4}{x + 1})^{^{'}}

8 t e^{2 t}

f (x) = \frac{5 - x e ^{x}}{x + e ^{x}}

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

What is the integral of (4x^3-6x^4)/(2x^6) ?
The integral of (4x^3-6x^4)/(2x^6) is (3x-1)/(x^2)+C