What is the integral of (3x^3-4x-2)/(x^3) ?

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

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integral of (3x^3-4x-2)/(x^3)

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

3 x + \frac{4}{x} + \frac{1}{x ^{2}} + C

Solution steps

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

Expand

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

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

\int 3 d x = 3 x

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

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

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

Simplify

= 3 x + \frac{4}{x} + \frac{1}{x ^{2}}

Add a constant to the solution

= 3 x + \frac{4}{x} + \frac{1}{x ^{2}} + C

\frac{( x + 5 )}{( x ^{2} + 5 x + 6 )}

y^{^{''}} - 2 y^{^{'}} - 3 y = - 9 t e - t

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

y^{^{'''}} - 3 y^{^{''}} + 3 y^{^{'}} - y = x - 2 e^{x}

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