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

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

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integral of (x^3+2x-6)/(x^3)

\int \frac{x ^{3} + 2 x - 6}{x ^{3}} d x

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

Solution steps

\int \frac{x ^{3} + 2 x - 6}{x ^{3}} d x

Expand

\frac{x ^{3} + 2 x - 6}{x ^{3}} : 1 + \frac{2}{x ^{2}} - \frac{6}{x ^{3}}

= \int 1 + \frac{2}{x ^{2}} - \frac{6}{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

= \int 1 d x + \int \frac{2}{x ^{2}} d x - \int \frac{6}{x ^{3}} d x

\int 1 d x = x

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

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

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

Simplify

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

Add a constant to the solution

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

\frac{d}{d x} (\frac{1}{4 x ^{\frac{3}{2}}})

k = 1 \sum \infty k x^{k - 1}

y^{^{''}} - 9 y^{^{'}} + 18 y = - 4 cos (3 t)

f (x) = \frac{14}{1 + e ^{- x}}

\int e^{x} 5 sin (2 x) d x

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