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

The integral of (3-4x^3+6x^6)/(x^6) is -3/(5x^5)+2/(x^2)+6x+C

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

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

- \frac{3}{5 x ^{5}} + \frac{2}{x ^{2}} + 6 x + C

Solution steps

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

Expand

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

= \int \frac{3}{x ^{6}} - \frac{4}{x ^{3}} + 6 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

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

\int \frac{3}{x ^{6}} d x = - \frac{3}{5 x ^{5}}

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

\int 6 d x = 6 x

= - \frac{3}{5 x ^{5}} - (- \frac{2}{x ^{2}}) + 6 x

Simplify

= - \frac{3}{5 x ^{5}} + \frac{2}{x ^{2}} + 6 x

Add a constant to the solution

= - \frac{3}{5 x ^{5}} + \frac{2}{x ^{2}} + 6 x + C

\frac{d y}{d x} = 1 - e^{x - y + 8}

- y^{^{'''}} + 2 y^{^{''}} + 3 y^{^{'}} + 3 = 0

h (x) = \frac{1}{x ^{10}}

\int \frac{1}{x ^{- 1} + 1} d x

\frac{d}{d x} (ln (x + 2 y))

What is the integral of (3-4x^3+6x^6)/(x^6) ?
The integral of (3-4x^3+6x^6)/(x^6) is -3/(5x^5)+2/(x^2)+6x+C