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

\int t^{2} - 4 t + 3 d t

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

\int_{1}^{8} x ln (x) d x

\int 8 x^{3} + 9 sin (x) d x

- \frac{12}{x ^{4}}

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