What is the integral of (1/(x^3)-x^3-1/8) ?

The integral of (1/(x^3)-x^3-1/8) is -1/(2x^2)-(x^4)/4-1/8 x+C

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integral of (1/(x^3)-x^3-1/8)

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

- \frac{1}{2 x ^{2}} - \frac{x ^{4}}{4} - \frac{1}{8} x + C

Solution steps

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

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

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

\int \frac{1}{8} d x = \frac{1}{8} x

= - \frac{1}{2 x ^{2}} - \frac{x ^{4}}{4} - \frac{1}{8} x

Add a constant to the solution

= - \frac{1}{2 x ^{2}} - \frac{x ^{4}}{4} - \frac{1}{8} x + C

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

x^{3} \frac{d y}{d x} - x^{2} y - y^{3} = 0

\frac{d}{d x} (\frac{5}{1 - 9 x})

cos (x^{2} - 5)

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

What is the integral of (1/(x^3)-x^3-1/8) ?
The integral of (1/(x^3)-x^3-1/8) is -1/(2x^2)-(x^4)/4-1/8 x+C