Solve ∫ left(2x-4right)^2 wrt x

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

Evaluate

$$\frac{4\left(x-2\right)^{3}}{3}+С$$

Solution Steps

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(2x-4\right)^{2}.

$$\int 4x^{2}-16x+16\mathrm{d}x$$

Integrate the sum term by term.

$$\int 4x^{2}\mathrm{d}x+\int -16x\mathrm{d}x+\int 16\mathrm{d}x$$

Factor out the constant in each of the terms.

$$4\int x^{2}\mathrm{d}x-16\int x\mathrm{d}x+\int 16\mathrm{d}x$$

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}.

$$\frac{4x^{3}}{3}-16\int x\mathrm{d}x+\int 16\mathrm{d}x$$

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}. Multiply -16 times \frac{x^{2}}{2}.

$$\frac{4x^{3}}{3}-8x^{2}+\int 16\mathrm{d}x$$

Find the integral of 16 using the table of common integrals rule \int a\mathrm{d}x=ax.

$$\frac{4x^{3}}{3}-8x^{2}+16x$$

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.

$$\frac{4x^{3}}{3}-8x^{2}+16x+С$$

Differentiate w.r.t. x

$$4\left(x-2\right)^{2}$$