Solve ∫ (2x^2-1)(3x+5) wrt x

$$\int ( 2 x ^ { 2 } - 1 ) ( 3 x + 5 ) d x$$

Evaluate

$$\frac{3x^{4}}{2}+\frac{10x^{3}}{3}-\frac{3x^{2}}{2}-5x+С$$

Solution Steps

Use the distributive property to multiply 2x^{2}-1 by 3x+5.

$$\int 6x^{3}+10x^{2}-3x-5\mathrm{d}x$$

Integrate the sum term by term.

$$\int 6x^{3}\mathrm{d}x+\int 10x^{2}\mathrm{d}x+\int -3x\mathrm{d}x+\int -5\mathrm{d}x$$

Factor out the constant in each of the terms.

$$6\int x^{3}\mathrm{d}x+10\int x^{2}\mathrm{d}x-3\int x\mathrm{d}x+\int -5\mathrm{d}x$$

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

$$\frac{3x^{4}}{2}+10\int x^{2}\mathrm{d}x-3\int x\mathrm{d}x+\int -5\mathrm{d}x$$

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

$$\frac{3x^{4}}{2}+\frac{10x^{3}}{3}-3\int x\mathrm{d}x+\int -5\mathrm{d}x$$

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 -3 times \frac{x^{2}}{2}.

$$\frac{3x^{4}}{2}+\frac{10x^{3}}{3}-\frac{3x^{2}}{2}+\int -5\mathrm{d}x$$

Find the integral of -5 using the table of common integrals rule \int a\mathrm{d}x=ax.

$$\frac{3x^{4}}{2}+\frac{10x^{3}}{3}-\frac{3x^{2}}{2}-5x$$

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.

$$\frac{3x^{4}}{2}+\frac{10x^{3}}{3}-\frac{3x^{2}}{2}-5x+С$$

Differentiate w.r.t. x

$$\left(3x+5\right)\left(2x^{2}-1\right)$$