What is the integral of (4x^3-3x+2)^2(4x^2-1) ?

The integral of (4x^3-3x+2)^2(4x^2-1) is (64x^9)/9-16x^7+(32x^6)/3+12x^5-16x^4+(7x^3)/3+6x^2-4x+C

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integral of (4x^3-3x+2)^2(4x^2-1)

\int (4 x^{3} - 3 x + 2)^{2} (4 x^{2} - 1) d x

\frac{64 x ^{9}}{9} - 16 x^{7} + \frac{32 x ^{6}}{3} + 12 x^{5} - 16 x^{4} + \frac{7 x ^{3}}{3} + 6 x^{2} - 4 x + C

Solution steps

\int (4 x^{3} - 3 x + 2)^{2} (4 x^{2} - 1) d x

Expand

(4 x^{3} - 3 x + 2)^{2} (4 x^{2} - 1) : 64 x^{8} - 112 x^{6} + 64 x^{5} + 60 x^{4} - 64 x^{3} + 7 x^{2} + 12 x - 4

= \int 64 x^{8} - 112 x^{6} + 64 x^{5} + 60 x^{4} - 64 x^{3} + 7 x^{2} + 12 x - 4 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 64 x^{8} d x - \int 112 x^{6} d x + \int 64 x^{5} d x + \int 60 x^{4} d x - \int 64 x^{3} d x + \int 7 x^{2} d x + \int 12 x d x - \int 4 d x

\int 64 x^{8} d x = \frac{64 x ^{9}}{9}

\int 112 x^{6} d x = 16 x^{7}

\int 64 x^{5} d x = \frac{32 x ^{6}}{3}

\int 60 x^{4} d x = 12 x^{5}

\int 64 x^{3} d x = 16 x^{4}

\int 7 x^{2} d x = \frac{7 x ^{3}}{3}

\int 12 x d x = 6 x^{2}

\int 4 d x = 4 x

= \frac{64 x ^{9}}{9} - 16 x^{7} + \frac{32 x ^{6}}{3} + 12 x^{5} - 16 x^{4} + \frac{7 x ^{3}}{3} + 6 x^{2} - 4 x

Add a constant to the solution

= \frac{64 x ^{9}}{9} - 16 x^{7} + \frac{32 x ^{6}}{3} + 12 x^{5} - 16 x^{4} + \frac{7 x ^{3}}{3} + 6 x^{2} - 4 x + C

\frac{d}{d x} (x e^{- x} cos (x))

\int \frac{2 x - 3}{3 x ^{4} + 7 x ^{2}} d x

x \to 0 lim (sin (\frac{x}{x}))

\frac{d}{d x} (- 8 x^{2})

(6 x^{2} y^{2} - 4 y^{4}) d x + (2 x^{3} y - 4 x y^{3}) d y = 0

What is the integral of (4x^3-3x+2)^2(4x^2-1) ?
The integral of (4x^3-3x+2)^2(4x^2-1) is (64x^9)/9-16x^7+(32x^6)/3+12x^5-16x^4+(7x^3)/3+6x^2-4x+C