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

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

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

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

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

Solution steps

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

Expand

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

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

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

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

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

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

\int 2 d x = 2 x

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

Add a constant to the solution

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

\frac{1}{x ^{2} + 4}

\int (2 cos (2 x) - 3 sin (3 x)) d x

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

- e^{- x}

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

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