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

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

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

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

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

Solution steps

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

Expand

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

= \int x^{2} + 3 - \frac{2}{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

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

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

\int 3 d x = 3 x

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

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

Simplify

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

Add a constant to the solution

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

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