Solve ∫ (x^2+5)(2x-3)

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

Evaluate

$$\frac{x^{4}}{2}-x^{3}+5x^{2}-15x+С$$

Solution Steps

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

$$\int 2x^{3}-3x^{2}+10x-15\mathrm{d}x$$

Integrate the sum term by term.

$$\int 2x^{3}\mathrm{d}x+\int -3x^{2}\mathrm{d}x+\int 10x\mathrm{d}x+\int -15\mathrm{d}x$$

Factor out the constant in each of the terms.

$$2\int x^{3}\mathrm{d}x-3\int x^{2}\mathrm{d}x+10\int x\mathrm{d}x+\int -15\mathrm{d}x$$

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

$$\frac{x^{4}}{2}-3\int x^{2}\mathrm{d}x+10\int x\mathrm{d}x+\int -15\mathrm{d}x$$

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

$$\frac{x^{4}}{2}-x^{3}+10\int x\mathrm{d}x+\int -15\mathrm{d}x$$

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

$$\frac{x^{4}}{2}-x^{3}+5x^{2}+\int -15\mathrm{d}x$$

Find the integral of -15 using the table of common integrals rule \int a\mathrm{d}x=ax.

$$\frac{x^{4}}{2}-x^{3}+5x^{2}-15x$$

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.

$$\frac{x^{4}}{2}-x^{3}+5x^{2}-15x+С$$

Differentiate w.r.t. x

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