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

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

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

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

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

Solution steps

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

Expand

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

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

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

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

\int 9 x^{2} d x = 3 x^{3}

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

Add a constant to the solution

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

y = e^{- x}, y = 0, x = 0, x = ln (5)

x \to 3 lim (a x + 3)

y d x + (x^{2} - 4 x) d y = 0

n = 1 \sum \infty \frac{9 n}{n !}

(ln^{2} (x))^{^{'}}

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