Solve ∫ (-x^3+3)(x+4) wrt x

$$\int ( - x ^ { 3 } + 3 ) ( x + 4 ) d x$$

Evaluate

$$-\frac{x^{5}}{5}-x^{4}+\frac{3x^{2}}{2}+12x+С$$

Solution Steps

Use the distributive property to multiply -x^{3}+3 by x+4.

$$\int \left(-x^{3}\right)x+4\left(-x^{3}\right)+3x+12\mathrm{d}x$$

To multiply powers of the same base, add their exponents. Add 3 and 1 to get 4.

$$\int -x^{4}+4\left(-1\right)x^{3}+3x+12\mathrm{d}x$$

Multiply 4 and -1 to get -4.

$$\int -x^{4}-4x^{3}+3x+12\mathrm{d}x$$

Integrate the sum term by term.

$$\int -x^{4}\mathrm{d}x+\int -4x^{3}\mathrm{d}x+\int 3x\mathrm{d}x+\int 12\mathrm{d}x$$

Factor out the constant in each of the terms.

$$-\int x^{4}\mathrm{d}x-4\int x^{3}\mathrm{d}x+3\int x\mathrm{d}x+\int 12\mathrm{d}x$$

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 -1 times \frac{x^{5}}{5}.

$$-\frac{x^{5}}{5}-4\int x^{3}\mathrm{d}x+3\int x\mathrm{d}x+\int 12\mathrm{d}x$$

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

$$-\frac{x^{5}}{5}-x^{4}+3\int x\mathrm{d}x+\int 12\mathrm{d}x$$

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}.

$$-\frac{x^{5}}{5}-x^{4}+\frac{3x^{2}}{2}+\int 12\mathrm{d}x$$

Find the integral of 12 using the table of common integrals rule \int a\mathrm{d}x=ax.

$$-\frac{x^{5}}{5}-x^{4}+\frac{3x^{2}}{2}+12x$$

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.

$$-\frac{x^{5}}{5}-x^{4}+\frac{3x^{2}}{2}+12x+С$$

Differentiate w.r.t. x

$$\left(-x-4\right)\left(x^{3}-3\right)$$