Solve ∫ (2x^2+5)(x^2-3x) wrt x=

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

Evaluate

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

Solution Steps

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

$$\int 2x^{4}-6x^{3}+5x^{2}-15x\mathrm{d}x$$

Integrate the sum term by term.

$$\int 2x^{4}\mathrm{d}x+\int -6x^{3}\mathrm{d}x+\int 5x^{2}\mathrm{d}x+\int -15x\mathrm{d}x$$

Factor out the constant in each of the terms.

$$2\int x^{4}\mathrm{d}x-6\int x^{3}\mathrm{d}x+5\int x^{2}\mathrm{d}x-15\int x\mathrm{d}x$$

Since \int x^{k}\mathrm{d}x=\frac{x^{k+1}}{k+1} for k\neq -1, replace \int x^{4}\mathrm{d}x with \frac{x^{5}}{5}. Multiply 2 times \frac{x^{5}}{5}.

$$\frac{2x^{5}}{5}-6\int x^{3}\mathrm{d}x+5\int x^{2}\mathrm{d}x-15\int x\mathrm{d}x$$

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}.

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

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

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

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 -15 times \frac{x^{2}}{2}.

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

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.

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

Differentiate w.r.t. x

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