Solve ∫ (4x-x^2-2x)^2 wrt x

$$\int ( 4 x - x ^ { 2 } - 2 x ) ^ { 2 } d x$$

Evaluate

$$\frac{x^{5}}{5}-x^{4}+\frac{4x^{3}}{3}+С$$

Solution Steps

Combine 4x and -2x to get 2x.

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

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

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

To multiply powers of the same base, add their exponents. Add 1 and 2 to get 3.

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

To raise a power to another power, multiply the exponents. Multiply 2 and 2 to get 4.

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

Integrate the sum term by term.

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

Factor out the constant in each of the terms.

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

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

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

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

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

Since \int x^{k}\mathrm{d}x=\frac{x^{k+1}}{k+1} for k\neq -1, replace \int x^{4}\mathrm{d}x with \frac{x^{5}}{5}.

$$\frac{4x^{3}}{3}-x^{4}+\frac{x^{5}}{5}$$

Simplify.

$$\frac{x^{5}}{5}-x^{4}+\frac{4x^{3}}{3}$$

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.

$$\frac{x^{5}}{5}-x^{4}+\frac{4x^{3}}{3}+С$$

Differentiate w.r.t. x

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