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

The integral of x^2(x+1)^2 is (x^5)/5+(x^4)/2+(x^3)/3+C

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

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

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

Solution steps

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

Expand

x^{2} (x + 1)^{2} : x^{4} + 2 x^{3} + x^{2}

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

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

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

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

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

Add a constant to the solution

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

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

5 x^{2} y

x \to 0 lim (\frac{7 x ^{3} - x ^{2}}{x})

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

x \to 1 - lim (f (x))

What is the integral of x^2(x+1)^2 ?
The integral of x^2(x+1)^2 is (x^5)/5+(x^4)/2+(x^3)/3+C