What is the integral of 4sin(x)-1+8x^{-5} ?

The integral of 4sin(x)-1+8x^{-5} is -4cos(x)-x-2/(x^4)+C

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integral of 4sin(x)-1+8x^{-5}

\int 4 sin (x) - 1 + 8 x^{- 5} d x

- 4 cos (x) - x - \frac{2}{x ^{4}} + C

Solution steps

\int 4 sin (x) - 1 + 8 x^{- 5} d x

8 x^{- 5} = \frac{8}{x ^{5}}

= \int 4 sin (x) - 1 + \frac{8}{x ^{5}} 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 4 sin (x) d x - \int 1 d x + \int \frac{8}{x ^{5}} d x

\int 4 sin (x) d x = - 4 cos (x)

\int 1 d x = x

\int \frac{8}{x ^{5}} d x = - \frac{2}{x ^{4}}

= - 4 cos (x) - x - \frac{2}{x ^{4}}

Add a constant to the solution

= - 4 cos (x) - x - \frac{2}{x ^{4}} + C

f (- 3) = 20 x^{4}

\int 3 e^{4 x + e^{4 x}} d x

\frac{d}{d x} (\frac{3 - 3 x - x ^{2}}{x ^{2} - 5})

\int \frac{1}{3 - x} d x

\int_{1}^{l n (x)} \frac{1}{x} d x

What is the integral of 4sin(x)-1+8x^{-5} ?
The integral of 4sin(x)-1+8x^{-5} is -4cos(x)-x-2/(x^4)+C