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

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

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

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

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

Solution steps

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

Expand

(x^{2} - 2 x + 3)^{2} (x + 1) : x^{5} - 3 x^{4} + 6 x^{3} - 2 x^{2} - 3 x + 9

= \int x^{5} - 3 x^{4} + 6 x^{3} - 2 x^{2} - 3 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 x^{5} d x - \int 3 x^{4} d x + \int 6 x^{3} d x - \int 2 x^{2} d x - \int 3 x d x + \int 9 d x

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

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

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

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

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

\int 9 d x = 9 x

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

Add a constant to the solution

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

\int x^{3} e^{2 x} d x

n = 1 \sum \infty \frac{n ^{5}}{2 ^{n}}

F (x) = sin (2 x) - 2 x cos (2 x)

\int 2 y e^{x y} d y

(10, 6), (- 5, - 3)

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