What is the integral of 1/(x^6)+3/(x^2) ?

The integral of 1/(x^6)+3/(x^2) is -1/(5x^5)-3/x+C

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integral of 1/(x^6)+3/(x^2)

\int \frac{1}{x ^{6}} + \frac{3}{x ^{2}} d x

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

Solution steps

\int \frac{1}{x ^{6}} + \frac{3}{x ^{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 ^{6}} d x + \int \frac{3}{x ^{2}} d x

\int \frac{1}{x ^{6}} d x = - \frac{1}{5 x ^{5}}

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

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

Add a constant to the solution

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

2 x e^{x} + e^{x} x^{2}

y = e^{- 7 x^{2}}

1 0^{x}

\int_{0}^{π} e^{- s t} sin (t) d t

\frac{d}{d x} (\frac{cos ( x )}{2 + sin ( x )})

What is the integral of 1/(x^6)+3/(x^2) ?
The integral of 1/(x^6)+3/(x^2) is -1/(5x^5)-3/x+C