Solve ∫ x^3+64/x+4 wrt x

$$\int{ \frac{ { x }^{ 3 } +64 }{ x+4 } }d x$$

Evaluate

$$\frac{x^{3}}{3}-2x^{2}+16x+С$$

Solution Steps

Factor the expressions that are not already factored in \frac{x^{3}+64}{x+4}.

$$\int \frac{\left(x+4\right)\left(x^{2}-4x+16\right)}{x+4}\mathrm{d}x$$

Cancel out x+4 in both numerator and denominator.

$$\int x^{2}-4x+16\mathrm{d}x$$

Integrate the sum term by term.

$$\int x^{2}\mathrm{d}x+\int -4x\mathrm{d}x+\int 16\mathrm{d}x$$

Factor out the constant in each of the terms.

$$\int x^{2}\mathrm{d}x-4\int x\mathrm{d}x+\int 16\mathrm{d}x$$

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}.

$$\frac{x^{3}}{3}-4\int x\mathrm{d}x+\int 16\mathrm{d}x$$

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

$$\frac{x^{3}}{3}-2x^{2}+\int 16\mathrm{d}x$$

Find the integral of 16 using the table of common integrals rule \int a\mathrm{d}x=ax.

$$\frac{x^{3}}{3}-2x^{2}+16x$$

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.

$$16x-2x^{2}+\frac{x^{3}}{3}+С$$

Differentiate w.r.t. x

$$x^{2}-4x+16$$