Solve ∫ (2x+1)(2x-1) wrt x

$$\int ( 2 x + 1 ) ( 2 x - 1 ) d x$$

Evaluate

$$\frac{4x^{3}}{3}-x+С$$

Solution Steps

Consider \left(2x+1\right)\left(2x-1\right). Multiplication can be transformed into difference of squares using the rule: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}.

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

Expand \left(2x\right)^{2}.

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

Calculate 2 to the power of 2 and get 4.

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

Calculate 1 to the power of 2 and get 1.

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

Integrate the sum term by term.

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

Factor out the constant in each of the terms.

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

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

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

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

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

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.

$$\frac{4x^{3}}{3}-x+С$$

Differentiate w.r.t. x

$$4x^{2}-1$$