Solve ∫ left(4x-2right)^3 wrt x

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

Evaluate

$$\left(2x-1\right)^{4}+С$$

Solution Steps

Use binomial theorem \left(a-b\right)^{3}=a^{3}-3a^{2}b+3ab^{2}-b^{3} to expand \left(4x-2\right)^{3}.

$$\int 64x^{3}-96x^{2}+48x-8\mathrm{d}x$$

Integrate the sum term by term.

$$\int 64x^{3}\mathrm{d}x+\int -96x^{2}\mathrm{d}x+\int 48x\mathrm{d}x+\int -8\mathrm{d}x$$

Factor out the constant in each of the terms.

$$64\int x^{3}\mathrm{d}x-96\int x^{2}\mathrm{d}x+48\int x\mathrm{d}x+\int -8\mathrm{d}x$$

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

$$16x^{4}-96\int x^{2}\mathrm{d}x+48\int x\mathrm{d}x+\int -8\mathrm{d}x$$

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 -96 times \frac{x^{3}}{3}.

$$16x^{4}-32x^{3}+48\int x\mathrm{d}x+\int -8\mathrm{d}x$$

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 48 times \frac{x^{2}}{2}.

$$16x^{4}-32x^{3}+24x^{2}+\int -8\mathrm{d}x$$

Find the integral of -8 using the table of common integrals rule \int a\mathrm{d}x=ax.

$$16x^{4}-32x^{3}+24x^{2}-8x$$

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.

$$16x^{4}-32x^{3}+24x^{2}-8x+С$$

Differentiate w.r.t. x

$$8\left(2x-1\right)^{3}$$