What is the integral of x^{-4}+2x-6x^2+cos(x) ?

The integral of x^{-4}+2x-6x^2+cos(x) is -1/(3x^3)+x^2-2x^3+sin(x)+C

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integral of x^{-4}+2x-6x^2+cos(x)

\int x^{- 4} + 2 x - 6 x^{2} + cos (x) d x

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

Solution steps

\int x^{- 4} + 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^{- 4} d x + \int 2 x d x - \int 6 x^{2} d x + \int cos (x) d x

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

\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}{3 x ^{3}} + x^{2} - 2 x^{3} + sin (x)

Add a constant to the solution

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

\int x f (x^{2}) d x

\int \frac{1}{x ( 25 - x )} d x

\int \frac{sec ( t )}{2 sec ^{2} ( t ) - 1} d t

\int \frac{x ^{2} + 5 x + 7}{x ^{2} + 3 x + 4} d x

What is the integral of x^{-4}+2x-6x^2+cos(x) ?
The integral of x^{-4}+2x-6x^2+cos(x) is -1/(3x^3)+x^2-2x^3+sin(x)+C