What is the integral of x^{-6}+2x-6x^2+cos(x) ?

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

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integral of x^{-6}+2x-6x^2+cos(x)

\int x^{- 6} + 2 x - 6 x^{2} + cos (x) d x

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

Solution steps

\int x^{- 6} + 2 x - 6 x^{2} + cos (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 x^{- 6} d x + \int 2 x d x - \int 6 x^{2} d x + \int cos (x) d x

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

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

\int 6 x^{2} d x = 2 x^{3}

\int cos (x) d x = sin (x)

= - \frac{1}{5 x ^{5}} + x^{2} - 2 x^{3} + sin (x)

Add a constant to the solution

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

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

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

\int 1 + x^{- 2} d x

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

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