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

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

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

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

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

Solution steps

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

Expand

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

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

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

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

\int 9 d x = 9 x

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

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

Add a constant to the solution

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

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