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

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

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integral of x^2(x+1)

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

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

Solution steps

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

Expand

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

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

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

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

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

Add a constant to the solution

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

x \to - 3 lim (\frac{x + 3}{x + 4})

\int \frac{( 1 + cos ( x ) )}{sin ( x )} d x

\frac{\partial}{\partial x} (e^{x} + y + y^{4} + 6)

y^{^{''}} + \frac{3}{x} y^{^{'}} - \frac{8}{x ^{2}} y = 0

\int_{- 5}^{5} \frac{1}{x} d x

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