What is the integral of 1/(x^3)-x ?

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

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integral of 1/(x^3)-x

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

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

Solution steps

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

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

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

= - \frac{1}{2 x ^{2}} - \frac{x ^{2}}{2}

Add a constant to the solution

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

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

x \to 0 lim (2)

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

n = 0 \sum \infty (\frac{6}{5})^{n}

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

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