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

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

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integral of (4x^5+2x^4+3x^2-x+6)

\int (4 x^{5} + 2 x^{4} + 3 x^{2} - x + 6) d x

\frac{2 x ^{6}}{3} + \frac{2 x ^{5}}{5} + x^{3} - \frac{x ^{2}}{2} + 6 x + C

Solution steps

\int 4 x^{5} + 2 x^{4} + 3 x^{2} - 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 4 x^{5} d x + \int 2 x^{4} d x + \int 3 x^{2} d x - \int x d x + \int 6 d x

\int 4 x^{5} d x = \frac{2 x ^{6}}{3}

\int 2 x^{4} d x = \frac{2 x ^{5}}{5}

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

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

\int 6 d x = 6 x

= \frac{2 x ^{6}}{3} + \frac{2 x ^{5}}{5} + x^{3} - \frac{x ^{2}}{2} + 6 x

Add a constant to the solution

= \frac{2 x ^{6}}{3} + \frac{2 x ^{5}}{5} + x^{3} - \frac{x ^{2}}{2} + 6 x + C

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