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

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

integral of (x^4-6x^3+7/(x^2))

\int (x^{4} - 6 x^{3} + \frac{7}{x ^{2}}) d x

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

Solution steps

\int x^{4} - 6 x^{3} + \frac{7}{x ^{2}} 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^{4} d x - \int 6 x^{3} d x + \int \frac{7}{x ^{2}} d x

\int x^{4} d x = \frac{x ^{5}}{5}

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

\int \frac{7}{x ^{2}} d x = - \frac{7}{x}

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

Add a constant to the solution

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

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