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

The integral of x(x+3)^4 is (x^6)/6+(12x^5)/5+(27x^4)/2+36x^3+(81x^2)/2+C

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integral of x(x+3)^4

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

\frac{x ^{6}}{6} + \frac{12 x ^{5}}{5} + \frac{27 x ^{4}}{2} + 36 x^{3} + \frac{81 x ^{2}}{2} + C

Solution steps

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

Expand

x (x + 3)^{4} : x^{5} + 12 x^{4} + 54 x^{3} + 108 x^{2} + 81 x

= \int x^{5} + 12 x^{4} + 54 x^{3} + 108 x^{2} + 81 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 x^{5} d x + \int 12 x^{4} d x + \int 54 x^{3} d x + \int 108 x^{2} d x + \int 81 x d x

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

\int 12 x^{4} d x = \frac{12 x ^{5}}{5}

\int 54 x^{3} d x = \frac{27 x ^{4}}{2}

\int 108 x^{2} d x = 36 x^{3}

\int 81 x d x = \frac{81 x ^{2}}{2}

= \frac{x ^{6}}{6} + \frac{12 x ^{5}}{5} + \frac{27 x ^{4}}{2} + 36 x^{3} + \frac{81 x ^{2}}{2}

Add a constant to the solution

= \frac{x ^{6}}{6} + \frac{12 x ^{5}}{5} + \frac{27 x ^{4}}{2} + 36 x^{3} + \frac{81 x ^{2}}{2} + C

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What is the integral of x(x+3)^4 ?
The integral of x(x+3)^4 is (x^6)/6+(12x^5)/5+(27x^4)/2+36x^3+(81x^2)/2+C