Solve ∫ x(1-x)^2 wrt x

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

Evaluate

$$\frac{x^{4}}{4}-\frac{2x^{3}}{3}+\frac{x^{2}}{2}+С$$

Solution Steps

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(1-x\right)^{2}.

$$\int x\left(1-2x+x^{2}\right)\mathrm{d}x$$

Use the distributive property to multiply x by 1-2x+x^{2}.

$$\int x-2x^{2}+x^{3}\mathrm{d}x$$

Integrate the sum term by term.

$$\int x\mathrm{d}x+\int -2x^{2}\mathrm{d}x+\int x^{3}\mathrm{d}x$$

Factor out the constant in each of the terms.

$$\int x\mathrm{d}x-2\int x^{2}\mathrm{d}x+\int x^{3}\mathrm{d}x$$

Since \int x^{k}\mathrm{d}x=\frac{x^{k+1}}{k+1} for k\neq -1, replace \int x\mathrm{d}x with \frac{x^{2}}{2}.

$$\frac{x^{2}}{2}-2\int x^{2}\mathrm{d}x+\int x^{3}\mathrm{d}x$$

Since \int x^{k}\mathrm{d}x=\frac{x^{k+1}}{k+1} for k\neq -1, replace \int x^{2}\mathrm{d}x with \frac{x^{3}}{3}. Multiply -2 times \frac{x^{3}}{3}.

$$\frac{x^{2}}{2}-\frac{2x^{3}}{3}+\int x^{3}\mathrm{d}x$$

Since \int x^{k}\mathrm{d}x=\frac{x^{k+1}}{k+1} for k\neq -1, replace \int x^{3}\mathrm{d}x with \frac{x^{4}}{4}.

$$\frac{x^{2}}{2}-\frac{2x^{3}}{3}+\frac{x^{4}}{4}$$

Simplify.

$$\frac{x^{4}}{4}-\frac{2x^{3}}{3}+\frac{x^{2}}{2}$$

If F\left(x\right) is an antiderivative of f\left(x\right), then the set of all antiderivatives of f\left(x\right) is given by F\left(x\right)+C. Therefore, add the constant of integration C\in \mathrm{R} to the result.

$$\frac{x^{4}}{4}-\frac{2x^{3}}{3}+\frac{x^{2}}{2}+С$$

Differentiate w.r.t. x

$$x\left(x-1\right)^{2}$$