Solve ∫ x^2(1-x)+1/3left(1-xright)^3+x(1-x) wrt x

$$\int{ { x }^{ 2 } (1-x)+ \frac{ 1 }{ 3 } { \left(1-x \right) }^{ 3 } +x(1-x) }d x$$

Evaluate

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

Solution Steps

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

$$\int x^{2}-x^{3}+\frac{1}{3}\left(1-x\right)^{3}+x\left(1-x\right)\mathrm{d}x$$

Use binomial theorem \left(a-b\right)^{3}=a^{3}-3a^{2}b+3ab^{2}-b^{3} to expand \left(1-x\right)^{3}.

$$\int x^{2}-x^{3}+\frac{1}{3}\left(1-3x+3x^{2}-x^{3}\right)+x\left(1-x\right)\mathrm{d}x$$

Use the distributive property to multiply \frac{1}{3} by 1-3x+3x^{2}-x^{3}.

$$\int x^{2}-x^{3}+\frac{1}{3}-x+x^{2}-\frac{1}{3}x^{3}+x\left(1-x\right)\mathrm{d}x$$

Combine x^{2} and x^{2} to get 2x^{2}.

$$\int 2x^{2}-x^{3}+\frac{1}{3}-x-\frac{1}{3}x^{3}+x\left(1-x\right)\mathrm{d}x$$

Combine -x^{3} and -\frac{1}{3}x^{3} to get -\frac{4}{3}x^{3}.

$$\int 2x^{2}-\frac{4}{3}x^{3}+\frac{1}{3}-x+x\left(1-x\right)\mathrm{d}x$$

Use the distributive property to multiply x by 1-x.

$$\int 2x^{2}-\frac{4}{3}x^{3}+\frac{1}{3}-x+x-x^{2}\mathrm{d}x$$

Combine -x and x to get 0.

$$\int 2x^{2}-\frac{4}{3}x^{3}+\frac{1}{3}-x^{2}\mathrm{d}x$$

Combine 2x^{2} and -x^{2} to get x^{2}.

$$\int x^{2}-\frac{4}{3}x^{3}+\frac{1}{3}\mathrm{d}x$$

Integrate the sum term by term.

$$\int x^{2}\mathrm{d}x+\int -\frac{4x^{3}}{3}\mathrm{d}x+\int \frac{1}{3}\mathrm{d}x$$

Factor out the constant in each of the terms.

$$\int x^{2}\mathrm{d}x-\frac{4\int x^{3}\mathrm{d}x}{3}+\int \frac{1}{3}\mathrm{d}x$$

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}.

$$\frac{x^{3}}{3}-\frac{4\int x^{3}\mathrm{d}x}{3}+\int \frac{1}{3}\mathrm{d}x$$

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}. Multiply -\frac{4}{3} times \frac{x^{4}}{4}.

$$\frac{x^{3}}{3}-\frac{x^{4}}{3}+\int \frac{1}{3}\mathrm{d}x$$

Find the integral of \frac{1}{3} using the table of common integrals rule \int a\mathrm{d}x=ax.

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

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.

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

Differentiate w.r.t. x

$$-\frac{4x^{3}}{3}+x^{2}+\frac{1}{3}$$