What is the integral of (1/(x^{-3)}+x^2e^{3x^3}-8/x) ?

The integral of (1/(x^{-3)}+x^2e^{3x^3}-8/x) is (x^4)/4+1/9 e^{3x^3}-8ln|x|+C

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integral of (1/(x^{-3)}+x^2e^{3x^3}-8/x)

\int (\frac{1}{x ^{- 3}} + x^{2} e^{3 x^{3}} - \frac{8}{x}) d x

\frac{x ^{4}}{4} + \frac{1}{9} e^{3 x^{3}} - 8 ln ∣ x ∣ + C

Solution steps

\int \frac{1}{x ^{- 3}} + x^{2} e^{3 x^{3}} - \frac{8}{x} 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 \frac{1}{x ^{- 3}} d x + \int x^{2} e^{3 x^{3}} d x - \int \frac{8}{x} d x

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

\int x^{2} e^{3 x^{3}} d x = \frac{1}{9} e^{3 x^{3}}

\int \frac{8}{x} d x = 8 ln ∣ x ∣

= \frac{x ^{4}}{4} + \frac{1}{9} e^{3 x^{3}} - 8 ln ∣ x ∣

Add a constant to the solution

= \frac{x ^{4}}{4} + \frac{1}{9} e^{3 x^{3}} - 8 ln ∣ x ∣ + C

y^{^{'}} + 6 y = e^{- t} - 6

\int e^{4 x} - 5 sin (x) d x

f (x) = \frac{8}{x ^{2} + 4}

\int 3 x^{3} + 4 x^{2} - 5 x d x

x \to 0 lim (\frac{2 ( y + x ) - 2 y}{x})

What is the integral of (1/(x^{-3)}+x^2e^{3x^3}-8/x) ?
The integral of (1/(x^{-3)}+x^2e^{3x^3}-8/x) is (x^4)/4+1/9 e^{3x^3}-8ln|x|+C