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

The integral of (x^2-3x+2)^3(2x-3) is (x^8)/4-3x^7+(31x^6)/2-45x^5+(321x^4)/4-90x^3+62x^2-24x+C

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

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

\frac{x ^{8}}{4} - 3 x^{7} + \frac{31 x ^{6}}{2} - 45 x^{5} + \frac{321 x ^{4}}{4} - 90 x^{3} + 62 x^{2} - 24 x + C

Solution steps

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

Expand

(x^{2} - 3 x + 2)^{3} (2 x - 3) : 2 x^{7} - 21 x^{6} + 93 x^{5} - 225 x^{4} + 321 x^{3} - 270 x^{2} + 124 x - 24

= \int 2 x^{7} - 21 x^{6} + 93 x^{5} - 225 x^{4} + 321 x^{3} - 270 x^{2} + 124 x - 24 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 21 x^{6} d x + \int 93 x^{5} d x - \int 225 x^{4} d x + \int 321 x^{3} d x - \int 270 x^{2} d x + \int 124 x d x - \int 24 d x

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

\int 21 x^{6} d x = 3 x^{7}

\int 93 x^{5} d x = \frac{31 x ^{6}}{2}

\int 225 x^{4} d x = 45 x^{5}

\int 321 x^{3} d x = \frac{321 x ^{4}}{4}

\int 270 x^{2} d x = 90 x^{3}

\int 124 x d x = 62 x^{2}

\int 24 d x = 24 x

= \frac{x ^{8}}{4} - 3 x^{7} + \frac{31 x ^{6}}{2} - 45 x^{5} + \frac{321 x ^{4}}{4} - 90 x^{3} + 62 x^{2} - 24 x

Add a constant to the solution

= \frac{x ^{8}}{4} - 3 x^{7} + \frac{31 x ^{6}}{2} - 45 x^{5} + \frac{321 x ^{4}}{4} - 90 x^{3} + 62 x^{2} - 24 x + C

\int (2 y) d y

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

\int \frac{4 x ^{2} + 16 x + 16}{x ^{3} + 8 x ^{2} + 16 x} d x

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

\int e^{(\frac{- x ^{2}}{2})} d x

What is the integral of (x^2-3x+2)^3(2x-3) ?
The integral of (x^2-3x+2)^3(2x-3) is (x^8)/4-3x^7+(31x^6)/2-45x^5+(321x^4)/4-90x^3+62x^2-24x+C