What is the integral of (9x^6+7x^5-x^4)/(6x^3) ?

The integral of (9x^6+7x^5-x^4)/(6x^3) is 1/6 ((9x^4)/4+(7x^3)/3-(x^2)/2)+C

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integral of (9x^6+7x^5-x^4)/(6x^3)

\int \frac{9 x ^{6} + 7 x ^{5} - x ^{4}}{6 x ^{3}} d x

\frac{1}{6} (\frac{9 x ^{4}}{4} + \frac{7 x ^{3}}{3} - \frac{x ^{2}}{2}) + C

Solution steps

\int \frac{9 x ^{6} + 7 x ^{5} - x ^{4}}{6 x ^{3}} d x

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= \frac{1}{6} \cdot \int \frac{9 x ^{6} + 7 x ^{5} - x ^{4}}{x ^{3}} d x

Expand

\frac{9 x ^{6} + 7 x ^{5} - x ^{4}}{x ^{3}} : \frac{9 x ^{6}}{x ^{3}} + \frac{7 x ^{5}}{x ^{3}} - \frac{x ^{4}}{x ^{3}}

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}{6} (\int \frac{9 x ^{6}}{x ^{3}} d x + \int \frac{7 x ^{5}}{x ^{3}} d x - \int \frac{x ^{4}}{x ^{3}} d x)

\int \frac{9 x ^{6}}{x ^{3}} d x = \frac{9 x ^{4}}{4}

\int \frac{7 x ^{5}}{x ^{3}} d x = \frac{7 x ^{3}}{3}

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

= \frac{1}{6} (\frac{9 x ^{4}}{4} + \frac{7 x ^{3}}{3} - \frac{x ^{2}}{2})

Add a constant to the solution

= \frac{1}{6} (\frac{9 x ^{4}}{4} + \frac{7 x ^{3}}{3} - \frac{x ^{2}}{2}) + C

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What is the integral of (9x^6+7x^5-x^4)/(6x^3) ?
The integral of (9x^6+7x^5-x^4)/(6x^3) is 1/6 ((9x^4)/4+(7x^3)/3-(x^2)/2)+C