What is the integral of (1/(x^4)-x^4-1/2) ?

The integral of (1/(x^4)-x^4-1/2) is -1/(3x^3)-(x^5)/5-1/2 x+C

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integral of (1/(x^4)-x^4-1/2)

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

- \frac{1}{3 x ^{3}} - \frac{x ^{5}}{5} - \frac{1}{2} x + C

Solution steps

\int \frac{1}{x ^{4}} - x^{4} - \frac{1}{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 \frac{1}{x ^{4}} d x - \int x^{4} d x - \int \frac{1}{2} d x

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

\int x^{4} d x = \frac{x ^{5}}{5}

\int \frac{1}{2} d x = \frac{1}{2} x

= - \frac{1}{3 x ^{3}} - \frac{x ^{5}}{5} - \frac{1}{2} x

Add a constant to the solution

= - \frac{1}{3 x ^{3}} - \frac{x ^{5}}{5} - \frac{1}{2} x + C

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What is the integral of (1/(x^4)-x^4-1/2) ?
The integral of (1/(x^4)-x^4-1/2) is -1/(3x^3)-(x^5)/5-1/2 x+C