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

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

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

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

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

Solution steps

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

Expand

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

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

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

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

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

\int 4 d x = 4 x

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

Add a constant to the solution

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

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