Solve ∫ -2(11+5x)^3

$$\int - 2 ( 11 + 5 x ) ^ { 3 }$$

Evaluate

$$-\frac{\left(5x+11\right)^{4}}{10}+С$$

Solution Steps

Use binomial theorem \left(a+b\right)^{3}=a^{3}+3a^{2}b+3ab^{2}+b^{3} to expand \left(11+5x\right)^{3}.

$$\int -2\left(1331+1815x+825x^{2}+125x^{3}\right)\mathrm{d}x$$

Use the distributive property to multiply -2 by 1331+1815x+825x^{2}+125x^{3}.

$$\int -2662-3630x-1650x^{2}-250x^{3}\mathrm{d}x$$

Integrate the sum term by term.

$$\int -2662\mathrm{d}x+\int -3630x\mathrm{d}x+\int -1650x^{2}\mathrm{d}x+\int -250x^{3}\mathrm{d}x$$

Factor out the constant in each of the terms.

$$\int -2662\mathrm{d}x-3630\int x\mathrm{d}x-1650\int x^{2}\mathrm{d}x-250\int x^{3}\mathrm{d}x$$

Find the integral of -2662 using the table of common integrals rule \int a\mathrm{d}x=ax.

$$-2662x-3630\int x\mathrm{d}x-1650\int x^{2}\mathrm{d}x-250\int x^{3}\mathrm{d}x$$

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 -3630 times \frac{x^{2}}{2}.

$$-2662x-1815x^{2}-1650\int x^{2}\mathrm{d}x-250\int x^{3}\mathrm{d}x$$

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 -1650 times \frac{x^{3}}{3}.

$$-2662x-1815x^{2}-550x^{3}-250\int x^{3}\mathrm{d}x$$

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 -250 times \frac{x^{4}}{4}.

$$-2662x-1815x^{2}-550x^{3}-\frac{125x^{4}}{2}$$

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.

$$-2662x-1815x^{2}-550x^{3}-\frac{125x^{4}}{2}+С$$

Differentiate w.r.t. x

$$-2\left(5x+11\right)^{3}$$