What is the integral of (8x^5-6x^4-2x-10)/(4x^3) ?

The integral of (8x^5-6x^4-2x-10)/(4x^3) is 1/4 ((8x^3)/3-3x^2+2/x+5/(x^2))+C

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integral of (8x^5-6x^4-2x-10)/(4x^3)

\int \frac{8 x ^{5} - 6 x ^{4} - 2 x - 10}{4 x ^{3}} d x

\frac{1}{4} (\frac{8 x ^{3}}{3} - 3 x^{2} + \frac{2}{x} + \frac{5}{x ^{2}}) + C

Solution steps

\int \frac{8 x ^{5} - 6 x ^{4} - 2 x - 10}{4 x ^{3}} d x

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= \frac{1}{4} \cdot \int \frac{8 x ^{5} - 6 x ^{4} - 2 x - 10}{x ^{3}} d x

Expand

\frac{8 x ^{5} - 6 x ^{4} - 2 x - 10}{x ^{3}} : 8 x^{2} - 6 x - \frac{2}{x ^{2}} - \frac{10}{x ^{3}}

= \frac{1}{4} \cdot \int 8 x^{2} - 6 x - \frac{2}{x ^{2}} - \frac{10}{x ^{3}} 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

= \frac{1}{4} (\int 8 x^{2} d x - \int 6 x d x - \int \frac{2}{x ^{2}} d x - \int \frac{10}{x ^{3}} d x)

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

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

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

\int \frac{10}{x ^{3}} d x = - \frac{5}{x ^{2}}

= \frac{1}{4} (\frac{8 x ^{3}}{3} - 3 x^{2} - (- \frac{2}{x}) - (- \frac{5}{x ^{2}}))

Simplify

= \frac{1}{4} (\frac{8 x ^{3}}{3} - 3 x^{2} + \frac{2}{x} + \frac{5}{x ^{2}})

Add a constant to the solution

= \frac{1}{4} (\frac{8 x ^{3}}{3} - 3 x^{2} + \frac{2}{x} + \frac{5}{x ^{2}}) + C

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What is the integral of (8x^5-6x^4-2x-10)/(4x^3) ?
The integral of (8x^5-6x^4-2x-10)/(4x^3) is 1/4 ((8x^3)/3-3x^2+2/x+5/(x^2))+C