Solve ∫ (x+8)^2 wrt x

$$\int ( x + 8 ) ^ { 2 } d x$$

Evaluate

$$\frac{\left(x+8\right)^{3}}{3}+С$$

Solution Steps

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(x+8\right)^{2}.

$$\int x^{2}+16x+64\mathrm{d}x$$

Integrate the sum term by term.

$$\int x^{2}\mathrm{d}x+\int 16x\mathrm{d}x+\int 64\mathrm{d}x$$

Factor out the constant in each of the terms.

$$\int x^{2}\mathrm{d}x+16\int x\mathrm{d}x+\int 64\mathrm{d}x$$

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}.

$$\frac{x^{3}}{3}+16\int x\mathrm{d}x+\int 64\mathrm{d}x$$

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

$$\frac{x^{3}}{3}+8x^{2}+\int 64\mathrm{d}x$$

Find the integral of 64 using the table of common integrals rule \int a\mathrm{d}x=ax.

$$\frac{x^{3}}{3}+8x^{2}+64x$$

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.

$$\frac{x^{3}}{3}+8x^{2}+64x+С$$

Differentiate w.r.t. x

$$\left(x+8\right)^{2}$$