What is the integral of (x^3-4+5/(x^6)) ?

The integral of (x^3-4+5/(x^6)) is (x^4)/4-4x-1/(x^5)+C

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integral of (x^3-4+5/(x^6))

\int (x^{3} - 4 + \frac{5}{x ^{6}}) d x

\frac{x ^{4}}{4} - 4 x - \frac{1}{x ^{5}} + C

Solution steps

\int x^{3} - 4 + \frac{5}{x ^{6}} 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^{3} d x - \int 4 d x + \int \frac{5}{x ^{6}} d x

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

\int 4 d x = 4 x

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

= \frac{x ^{4}}{4} - 4 x - \frac{1}{x ^{5}}

Add a constant to the solution

= \frac{x ^{4}}{4} - 4 x - \frac{1}{x ^{5}} + C

What is the integral of (x^3-4+5/(x^6)) ?
The integral of (x^3-4+5/(x^6)) is (x^4)/4-4x-1/(x^5)+C