What is the integral of (1/(x^5)-x^5-1/9) ?

The integral of (1/(x^5)-x^5-1/9) is -1/(4x^4)-(x^6)/6-1/9 x+C

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integral of (1/(x^5)-x^5-1/9)

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

- \frac{1}{4 x ^{4}} - \frac{x ^{6}}{6} - \frac{1}{9} x + C

Solution steps

\int \frac{1}{x ^{5}} - x^{5} - \frac{1}{9} 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 ^{5}} d x - \int x^{5} d x - \int \frac{1}{9} d x

\int \frac{1}{x ^{5}} d x = - \frac{1}{4 x ^{4}}

\int x^{5} d x = \frac{x ^{6}}{6}

\int \frac{1}{9} d x = \frac{1}{9} x

= - \frac{1}{4 x ^{4}} - \frac{x ^{6}}{6} - \frac{1}{9} x

Add a constant to the solution

= - \frac{1}{4 x ^{4}} - \frac{x ^{6}}{6} - \frac{1}{9} x + C

x \to - 2 - lim (- 4 x - 8 + 3)

\frac{d}{d x} (tanh (1 + e^{4 x}))

x \to - \infty lim (3 x^{2} + 1)

y^{^{'}} + 2 sin (2 x) y = 2 sin (2 x)

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

What is the integral of (1/(x^5)-x^5-1/9) ?
The integral of (1/(x^5)-x^5-1/9) is -1/(4x^4)-(x^6)/6-1/9 x+C