Solve ∫ left(3x-1right)^2 wrt x

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

Evaluate

$$\frac{\left(3x-1\right)^{3}}{9}+С$$

Solution Steps

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

$$\int 9x^{2}-6x+1\mathrm{d}x$$

Integrate the sum term by term.

$$\int 9x^{2}\mathrm{d}x+\int -6x\mathrm{d}x+\int 1\mathrm{d}x$$

Factor out the constant in each of the terms.

$$9\int x^{2}\mathrm{d}x-6\int x\mathrm{d}x+\int 1\mathrm{d}x$$

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

$$3x^{3}-6\int x\mathrm{d}x+\int 1\mathrm{d}x$$

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 -6 times \frac{x^{2}}{2}.

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

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

$$3x^{3}-3x^{2}+x$$

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.

$$3x^{3}-3x^{2}+x+С$$

Differentiate w.r.t. x

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