What is the integral of x^3-3x^{-4}-2 ?

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

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integral of x^3-3x^{-4}-2

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

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

Solution steps

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

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

\int 3 x^{- 4} d x = - \frac{1}{x ^{3}}

\int 2 d x = 2 x

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

Simplify

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

Add a constant to the solution

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

\int x^{2} lo g_{e} (x) d x

\frac{d y}{d t} = y + 1

\int_{0}^{7} 49 - x^{2} d x

\frac{\partial}{\partial x} ((x - 2) (1 - e^{- x}))

\frac{1}{4} y^{^{''}} + 4 t = 2 tan (4 t) - \frac{1}{3} e^{t}

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