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

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

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

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

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

Solution steps

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

Simplify

\frac{x ^{4} - 6 x ^{3} - 7 x}{x} : x^{3} - 6 x^{2} - 7

= \int x^{3} - 6 x^{2} - 7 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^{3} d x - \int 6 x^{2} d x - \int 7 d x

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

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

\int 7 d x = 7 x

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

Add a constant to the solution

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

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