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

The integral of x^2(x^3+1)^3 is (x^{12}}{12}+\frac{x^9)/3+(x^6)/2+(x^3)/3+C

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

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

\frac{x ^{12}}{12} + \frac{x ^{9}}{3} + \frac{x ^{6}}{2} + \frac{x ^{3}}{3} + C

Solution steps

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

Expand

x^{2} (x^{3} + 1)^{3} : x^{11} + 3 x^{8} + 3 x^{5} + x^{2}

= \int x^{11} + 3 x^{8} + 3 x^{5} + 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^{11} d x + \int 3 x^{8} d x + \int 3 x^{5} d x + \int x^{2} d x

\int x^{11} d x = \frac{x ^{12}}{12}

\int 3 x^{8} d x = \frac{x ^{9}}{3}

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

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

= \frac{x ^{12}}{12} + \frac{x ^{9}}{3} + \frac{x ^{6}}{2} + \frac{x ^{3}}{3}

Add a constant to the solution

= \frac{x ^{12}}{12} + \frac{x ^{9}}{3} + \frac{x ^{6}}{2} + \frac{x ^{3}}{3} + C

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

\frac{d y}{d x} = \frac{x ^{2} - y ^{2}}{x y}, y (2) = 2

4 x, x^{2} - 5

\int se x (x)

f (x) = x^{2} - 4, 1, 3

What is the integral of x^2(x^3+1)^3 ?
The integral of x^2(x^3+1)^3 is (x^{12}}{12}+\frac{x^9)/3+(x^6)/2+(x^3)/3+C