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

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

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integral of x(x+1)^2

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

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

Solution steps

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

Expand

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

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

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

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

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

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

Add a constant to the solution

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

\frac{d y}{d x} + \frac{1}{x + 1} y = x^{- 2}

\int \frac{1}{x x ^{2} + 16} d x

\int 4 x e^{3 x} d x

e^{x} + e^{- x}

\frac{\partial}{\partial x} (cos (x + e^{y}))

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