What is the integral of (1-x^2-3x^5) ?

The integral of (1-x^2-3x^5) is x-(x^3)/3-(x^6)/2+C

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integral of (1-x^2-3x^5)

\int (1 - x^{2} - 3 x^{5}) d x

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

Solution steps

\int 1 - x^{2} - 3 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 1 d x - \int x^{2} d x - \int 3 x^{5} d x

\int 1 d x = x

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

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

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

Add a constant to the solution

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

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What is the integral of (1-x^2-3x^5) ?
The integral of (1-x^2-3x^5) is x-(x^3)/3-(x^6)/2+C