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

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

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

\int (\frac{8 x ^{4} - 4 x ^{3} - 7 x}{x}) d x

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

Solution steps

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

Simplify

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

= \int 8 x^{3} - 4 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 4 x^{2} d x - \int 7 d x

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

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

\int 7 d x = 7 x

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

Add a constant to the solution

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

y \frac{d y}{d x} = e^{x}

\frac{\partial}{\partial x} (cos (x^{2}) \cdot sin (z \cdot y))

\frac{\partial}{\partial x} (17 - 2 x^{2} - y^{2})

\frac{\partial}{\partial x} (3 x^{2} y^{2} - 4)

x^{3} + 4 x

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