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

The integral of (2x^2+4x-3)^2(x+1) is (2x^6)/3+4x^5+5x^4-(20x^3)/3-(15x^2)/2+9x+C

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

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

\frac{2 x ^{6}}{3} + 4 x^{5} + 5 x^{4} - \frac{20 x ^{3}}{3} - \frac{15 x ^{2}}{2} + 9 x + C

Solution steps

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

Expand

(2 x^{2} + 4 x - 3)^{2} (x + 1) : 4 x^{5} + 20 x^{4} + 20 x^{3} - 20 x^{2} - 15 x + 9

= \int 4 x^{5} + 20 x^{4} + 20 x^{3} - 20 x^{2} - 15 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 4 x^{5} d x + \int 20 x^{4} d x + \int 20 x^{3} d x - \int 20 x^{2} d x - \int 15 x d x + \int 9 d x

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

\int 20 x^{4} d x = 4 x^{5}

\int 20 x^{3} d x = 5 x^{4}

\int 20 x^{2} d x = \frac{20 x ^{3}}{3}

\int 15 x d x = \frac{15 x ^{2}}{2}

\int 9 d x = 9 x

= \frac{2 x ^{6}}{3} + 4 x^{5} + 5 x^{4} - \frac{20 x ^{3}}{3} - \frac{15 x ^{2}}{2} + 9 x

Add a constant to the solution

= \frac{2 x ^{6}}{3} + 4 x^{5} + 5 x^{4} - \frac{20 x ^{3}}{3} - \frac{15 x ^{2}}{2} + 9 x + C

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What is the integral of (2x^2+4x-3)^2(x+1) ?
The integral of (2x^2+4x-3)^2(x+1) is (2x^6)/3+4x^5+5x^4-(20x^3)/3-(15x^2)/2+9x+C