Solve ∫ (x^3+2)^23x^2 wrt x

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

Evaluate

$$\frac{x^{9}}{3}+2x^{6}+4x^{3}+С$$

Solution Steps

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

$$\int \left(\left(x^{3}\right)^{2}+4x^{3}+4\right)\times 3x^{2}\mathrm{d}x$$

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

$$\int \left(x^{6}+4x^{3}+4\right)\times 3x^{2}\mathrm{d}x$$

Use the distributive property to multiply x^{6}+4x^{3}+4 by 3.

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

Use the distributive property to multiply 3x^{6}+12x^{3}+12 by x^{2}.

$$\int 3x^{8}+12x^{5}+12x^{2}\mathrm{d}x$$

Integrate the sum term by term.

$$\int 3x^{8}\mathrm{d}x+\int 12x^{5}\mathrm{d}x+\int 12x^{2}\mathrm{d}x$$

Factor out the constant in each of the terms.

$$3\int x^{8}\mathrm{d}x+12\int x^{5}\mathrm{d}x+12\int x^{2}\mathrm{d}x$$

Since \int x^{k}\mathrm{d}x=\frac{x^{k+1}}{k+1} for k\neq -1, replace \int x^{8}\mathrm{d}x with \frac{x^{9}}{9}. Multiply 3 times \frac{x^{9}}{9}.

$$\frac{x^{9}}{3}+12\int x^{5}\mathrm{d}x+12\int x^{2}\mathrm{d}x$$

Since \int x^{k}\mathrm{d}x=\frac{x^{k+1}}{k+1} for k\neq -1, replace \int x^{5}\mathrm{d}x with \frac{x^{6}}{6}. Multiply 12 times \frac{x^{6}}{6}.

$$\frac{x^{9}}{3}+2x^{6}+12\int x^{2}\mathrm{d}x$$

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 12 times \frac{x^{3}}{3}.

$$\frac{x^{9}}{3}+2x^{6}+4x^{3}$$

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.

$$4x^{3}+2x^{6}+\frac{x^{9}}{3}+С$$

Differentiate w.r.t. x

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