Solve ∫ (2x-1)(3x+4) wrt x=

$$\int ( 2 x - 1 ) ( 3 x + 4 ) d x =$$

Evaluate

$$2x^{3}+\frac{5x^{2}}{2}-4x+С$$

Solution Steps

Apply the distributive property by multiplying each term of 2x-1 by each term of 3x+4.

$$\int 6x^{2}+8x-3x-4\mathrm{d}x$$

Combine 8x and -3x to get 5x.

$$\int 6x^{2}+5x-4\mathrm{d}x$$

Integrate the sum term by term.

$$\int 6x^{2}\mathrm{d}x+\int 5x\mathrm{d}x+\int -4\mathrm{d}x$$

Factor out the constant in each of the terms.

$$6\int x^{2}\mathrm{d}x+5\int x\mathrm{d}x+\int -4\mathrm{d}x$$

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

$$2x^{3}+5\int x\mathrm{d}x+\int -4\mathrm{d}x$$

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

$$2x^{3}+\frac{5x^{2}}{2}+\int -4\mathrm{d}x$$

Find the integral of -4 using the table of common integrals rule \int a\mathrm{d}x=ax.

$$2x^{3}+\frac{5x^{2}}{2}-4x$$

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.

$$2x^{3}+\frac{5x^{2}}{2}-4x+С$$

Differentiate w.r.t. x

$$\left(2x-1\right)\left(3x+4\right)$$