What is the integral of (4x^6+3x^5-8)/(x^5) ?

The integral of (4x^6+3x^5-8)/(x^5) is 2x^2+3x+2/(x^4)+C

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integral of (4x^6+3x^5-8)/(x^5)

\int \frac{4 x ^{6} + 3 x ^{5} - 8}{x ^{5}} d x

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

Solution steps

\int \frac{4 x ^{6} + 3 x ^{5} - 8}{x ^{5}} d x

Expand

\frac{4 x ^{6} + 3 x ^{5} - 8}{x ^{5}} : 4 x + 3 - \frac{8}{x ^{5}}

= \int 4 x + 3 - \frac{8}{x ^{5}} 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 4 x d x + \int 3 d x - \int \frac{8}{x ^{5}} d x

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

\int 3 d x = 3 x

\int \frac{8}{x ^{5}} d x = - \frac{2}{x ^{4}}

= 2 x^{2} + 3 x - (- \frac{2}{x ^{4}})

Simplify

= 2 x^{2} + 3 x + \frac{2}{x ^{4}}

Add a constant to the solution

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

\frac{d y}{d x} = \frac{( e ^{x} - 1 )}{e ^{y}}

\frac{\partial}{\partial y} (1 + ln (x) + \frac{y}{x})

\int e^{4 x^{2}} d x

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

\frac{\partial}{\partial y} (\frac{18 x y}{x + y})

What is the integral of (4x^6+3x^5-8)/(x^5) ?
The integral of (4x^6+3x^5-8)/(x^5) is 2x^2+3x+2/(x^4)+C