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

The integral of (3x^5-6/(x^4)) is (x^6)/2+2/(x^3)+C

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

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

\frac{x ^{6}}{2} + \frac{2}{x ^{3}} + C

Solution steps

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

\int 3 x^{5} d x = \frac{x ^{6}}{2}

\int \frac{6}{x ^{4}} d x = - \frac{2}{x ^{3}}

= \frac{x ^{6}}{2} - (- \frac{2}{x ^{3}})

Simplify

= \frac{x ^{6}}{2} + \frac{2}{x ^{3}}

Add a constant to the solution

= \frac{x ^{6}}{2} + \frac{2}{x ^{3}} + C

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What is the integral of (3x^5-6/(x^4)) ?
The integral of (3x^5-6/(x^4)) is (x^6)/2+2/(x^3)+C