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

The integral of (2x-1)(x^2-x+3)^3 is (x^8)/4-x^7+(9x^6)/2-10x^5+(91x^4)/4-30x^3+(81x^2)/2-27x+C

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

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

\frac{x ^{8}}{4} - x^{7} + \frac{9 x ^{6}}{2} - 10 x^{5} + \frac{91 x ^{4}}{4} - 30 x^{3} + \frac{81 x ^{2}}{2} - 27 x + C

Solution steps

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

Expand

(2 x - 1) (x^{2} - x + 3)^{3} : 2 x^{7} - 7 x^{6} + 27 x^{5} - 50 x^{4} + 91 x^{3} - 90 x^{2} + 81 x - 27

= \int 2 x^{7} - 7 x^{6} + 27 x^{5} - 50 x^{4} + 91 x^{3} - 90 x^{2} + 81 x - 27 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 2 x^{7} d x - \int 7 x^{6} d x + \int 27 x^{5} d x - \int 50 x^{4} d x + \int 91 x^{3} d x - \int 90 x^{2} d x + \int 81 x d x - \int 27 d x

\int 2 x^{7} d x = \frac{x ^{8}}{4}

\int 7 x^{6} d x = x^{7}

\int 27 x^{5} d x = \frac{9 x ^{6}}{2}

\int 50 x^{4} d x = 10 x^{5}

\int 91 x^{3} d x = \frac{91 x ^{4}}{4}

\int 90 x^{2} d x = 30 x^{3}

\int 81 x d x = \frac{81 x ^{2}}{2}

\int 27 d x = 27 x

= \frac{x ^{8}}{4} - x^{7} + \frac{9 x ^{6}}{2} - 10 x^{5} + \frac{91 x ^{4}}{4} - 30 x^{3} + \frac{81 x ^{2}}{2} - 27 x

Add a constant to the solution

= \frac{x ^{8}}{4} - x^{7} + \frac{9 x ^{6}}{2} - 10 x^{5} + \frac{91 x ^{4}}{4} - 30 x^{3} + \frac{81 x ^{2}}{2} - 27 x + C

\int x - a d x

\int \frac{u}{u} d u

\int (x^{\frac{3}{2}} + 3)^{4} x d x

\int x^{3} 1 + 3 x^{2} d x

\int \frac{x ^{2} - 6}{x ^{2} - 9} d x

What is the integral of (2x-1)(x^2-x+3)^3 ?
The integral of (2x-1)(x^2-x+3)^3 is (x^8)/4-x^7+(9x^6)/2-10x^5+(91x^4)/4-30x^3+(81x^2)/2-27x+C