What is the integral of 3x^5-6x^2+x-9 ?

The integral of 3x^5-6x^2+x-9 is (x^6)/2-2x^3+(x^2)/2-9x+C

Popular >

integral of 3x^5-6x^2+x-9

\int 3 x^{5} - 6 x^{2} + x - 9 d x

\frac{x ^{6}}{2} - 2 x^{3} + \frac{x ^{2}}{2} - 9 x + C

Solution steps

\int 3 x^{5} - 6 x^{2} + x - 9 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 6 x^{2} d x + \int x d x - \int 9 d x

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

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

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

\int 9 d x = 9 x

= \frac{x ^{6}}{2} - 2 x^{3} + \frac{x ^{2}}{2} - 9 x

Add a constant to the solution

= \frac{x ^{6}}{2} - 2 x^{3} + \frac{x ^{2}}{2} - 9 x + C

\int 3 x^{2} x^{3} + 4 d x

\int \frac{2 x ^{2}}{a ^{2} + x ^{2}} d x

\int 2^{3 w} d w

\int (x^{- 3} - 6 x^{- 2} + 10) d x

\int \frac{x ^{2} + x - 2}{x - 1} d x

What is the integral of 3x^5-6x^2+x-9 ?
The integral of 3x^5-6x^2+x-9 is (x^6)/2-2x^3+(x^2)/2-9x+C