What is the integral of x^3-3x^{-4}+5 ?

The integral of x^3-3x^{-4}+5 is (x^4)/4+1/(x^3)+5x+C

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integral of x^3-3x^{-4}+5

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

\frac{x ^{4}}{4} + \frac{1}{x ^{3}} + 5 x + C

Solution steps

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

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

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

\int 5 d x = 5 x

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

Simplify

= \frac{x ^{4}}{4} + \frac{1}{x ^{3}} + 5 x

Add a constant to the solution

= \frac{x ^{4}}{4} + \frac{1}{x ^{3}} + 5 x + C

\frac{d}{d x} (xx)

\int_{- 1}^{1} 6 - 6 x^{2} d x

\frac{d y}{d x} = 10 y e^{x + 6}

\frac{d}{d x} (\frac{e ^{x}}{5 - x})

f (x) = ln (ln (x^{3} e^{- 3 x}))

What is the integral of x^3-3x^{-4}+5 ?
The integral of x^3-3x^{-4}+5 is (x^4)/4+1/(x^3)+5x+C