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

The integral of (5x^3+6x^3-5) is (11x^4)/4-5x+C

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

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

\frac{11 x ^{4}}{4} - 5 x + C

Solution steps

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

Simplify

= \int 11 x^{3} - 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 11 x^{3} d x - \int 5 d x

\int 11 x^{3} d x = \frac{11 x ^{4}}{4}

\int 5 d x = 5 x

= \frac{11 x ^{4}}{4} - 5 x

Add a constant to the solution

= \frac{11 x ^{4}}{4} - 5 x + C

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What is the integral of (5x^3+6x^3-5) ?
The integral of (5x^3+6x^3-5) is (11x^4)/4-5x+C