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

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

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

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

- \frac{7}{5 x ^{5}} + \frac{2}{x ^{2}} + 6 x + C

Solution steps

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

Expand

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

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

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

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

\int 6 d x = 6 x

= - \frac{7}{5 x ^{5}} - (- \frac{2}{x ^{2}}) + 6 x

Simplify

= - \frac{7}{5 x ^{5}} + \frac{2}{x ^{2}} + 6 x

Add a constant to the solution

= - \frac{7}{5 x ^{5}} + \frac{2}{x ^{2}} + 6 x + C

3 y^{^{''}} - 8 y^{^{'}} - 3 y = 0, y (0) = - 9, y^{^{'}} (0) = 8

\int_{1}^{3} (\frac{3 x ^{2} - 7}{x ^{3}}) d x

\int t^{3} (7 + t^{4})^{3} d t

x \to 1 lim ((x)^{\frac{1}{x - 1}})

\int_{0}^{4} 3 x d x

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