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

f (x) = 2 x^{4} + 3 x - 3, (2, 35)

9 a^{2} + 8 a - 56

t = 0 \sum \infty \frac{1}{t ^{0.51}}

\frac{d}{d x} (tan (10 x))

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