What is the integral of (8x^4+18x^3-7x)/x ?

The integral of (8x^4+18x^3-7x)/x is 2x^4+6x^3-7x+C

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integral of (8x^4+18x^3-7x)/x

\int \frac{8 x ^{4} + 18 x ^{3} - 7 x}{x} d x

2 x^{4} + 6 x^{3} - 7 x + C

Solution steps

\int \frac{8 x ^{4} + 18 x ^{3} - 7 x}{x} d x

Simplify

\frac{8 x ^{4} + 18 x ^{3} - 7 x}{x} : 8 x^{3} + 18 x^{2} - 7

= \int 8 x^{3} + 18 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 8 x^{3} d x + \int 18 x^{2} d x - \int 7 d x

\int 8 x^{3} d x = 2 x^{4}

\int 18 x^{2} d x = 6 x^{3}

\int 7 d x = 7 x

= 2 x^{4} + 6 x^{3} - 7 x

Add a constant to the solution

= 2 x^{4} + 6 x^{3} - 7 x + C

- 2 x^{2}

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What is the integral of (8x^4+18x^3-7x)/x ?
The integral of (8x^4+18x^3-7x)/x is 2x^4+6x^3-7x+C