Solve ∫ (x^3+2)^3 wrt x

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

Evaluate

$$\frac{x^{10}}{10}+\frac{6x^{7}}{7}+3x^{4}+8x+С$$

Solution Steps

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

$$\int \left(x^{3}\right)^{3}+6\left(x^{3}\right)^{2}+12x^{3}+8\mathrm{d}x$$

To raise a power to another power, multiply the exponents. Multiply 3 and 3 to get 9.

$$\int x^{9}+6\left(x^{3}\right)^{2}+12x^{3}+8\mathrm{d}x$$

To raise a power to another power, multiply the exponents. Multiply 3 and 2 to get 6.

$$\int x^{9}+6x^{6}+12x^{3}+8\mathrm{d}x$$

Integrate the sum term by term.

$$\int x^{9}\mathrm{d}x+\int 6x^{6}\mathrm{d}x+\int 12x^{3}\mathrm{d}x+\int 8\mathrm{d}x$$

Factor out the constant in each of the terms.

$$\int x^{9}\mathrm{d}x+6\int x^{6}\mathrm{d}x+12\int x^{3}\mathrm{d}x+\int 8\mathrm{d}x$$

Since \int x^{k}\mathrm{d}x=\frac{x^{k+1}}{k+1} for k\neq -1, replace \int x^{9}\mathrm{d}x with \frac{x^{10}}{10}.

$$\frac{x^{10}}{10}+6\int x^{6}\mathrm{d}x+12\int x^{3}\mathrm{d}x+\int 8\mathrm{d}x$$

Since \int x^{k}\mathrm{d}x=\frac{x^{k+1}}{k+1} for k\neq -1, replace \int x^{6}\mathrm{d}x with \frac{x^{7}}{7}. Multiply 6 times \frac{x^{7}}{7}.

$$\frac{x^{10}}{10}+\frac{6x^{7}}{7}+12\int x^{3}\mathrm{d}x+\int 8\mathrm{d}x$$

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

$$\frac{x^{10}}{10}+\frac{6x^{7}}{7}+3x^{4}+\int 8\mathrm{d}x$$

Find the integral of 8 using the table of common integrals rule \int a\mathrm{d}x=ax.

$$\frac{x^{10}}{10}+\frac{6x^{7}}{7}+3x^{4}+8x$$

Simplify.

$$8x+3x^{4}+\frac{6x^{7}}{7}+\frac{x^{10}}{10}$$

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.

$$8x+3x^{4}+\frac{6x^{7}}{7}+\frac{x^{10}}{10}+С$$

Differentiate w.r.t. x

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