What is the integral of (7-6x^3+4x^6)/(x^6) ?

The integral of (7-6x^3+4x^6)/(x^6) is -7/(5x^5)+3/(x^2)+4x+C

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integral of (7-6x^3+4x^6)/(x^6)

\int \frac{7 - 6 x ^{3} + 4 x ^{6}}{x ^{6}} d x

- \frac{7}{5 x ^{5}} + \frac{3}{x ^{2}} + 4 x + C

Solution steps

\int \frac{7 - 6 x ^{3} + 4 x ^{6}}{x ^{6}} d x

Expand

\frac{7 - 6 x ^{3} + 4 x ^{6}}{x ^{6}} : \frac{7}{x ^{6}} - \frac{6}{x ^{3}} + 4

= \int \frac{7}{x ^{6}} - \frac{6}{x ^{3}} + 4 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{7}{x ^{6}} d x - \int \frac{6}{x ^{3}} d x + \int 4 d x

\int \frac{7}{x ^{6}} d x = - \frac{7}{5 x ^{5}}

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

\int 4 d x = 4 x

= - \frac{7}{5 x ^{5}} - (- \frac{3}{x ^{2}}) + 4 x

Simplify

= - \frac{7}{5 x ^{5}} + \frac{3}{x ^{2}} + 4 x

Add a constant to the solution

= - \frac{7}{5 x ^{5}} + \frac{3}{x ^{2}} + 4 x + C

\frac{d}{d x} (arctan (\frac{1}{3 x}))

n = 0 \sum \infty 3 (\frac{1}{2})^{n}

\frac{d}{d r} (\frac{r}{r ^{2} + 5})

\frac{\partial}{\partial y} (x^{2} + y)

f (x) = \frac{3 x + 1}{x ^{2} + 1}, (1, 2)

What is the integral of (7-6x^3+4x^6)/(x^6) ?
The integral of (7-6x^3+4x^6)/(x^6) is -7/(5x^5)+3/(x^2)+4x+C