Solve ∫ 2xleft(x^2-3right)^2 wrt x

$$\int{ 2x { \left( { x }^{ 2 } -3 \right) }^{ 2 } }d x$$

Evaluate

$$\frac{x^{6}}{3}-3x^{4}+9x^{2}+С$$

Solution Steps

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

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

To raise a power to another power, multiply the exponents. Multiply 2 and 2 to get 4.

$$\int 2x\left(x^{4}-6x^{2}+9\right)\mathrm{d}x$$

Use the distributive property to multiply 2x by x^{4}-6x^{2}+9.

$$\int 2x^{5}-12x^{3}+18x\mathrm{d}x$$

Integrate the sum term by term.

$$\int 2x^{5}\mathrm{d}x+\int -12x^{3}\mathrm{d}x+\int 18x\mathrm{d}x$$

Factor out the constant in each of the terms.

$$2\int x^{5}\mathrm{d}x-12\int x^{3}\mathrm{d}x+18\int x\mathrm{d}x$$

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 2 times \frac{x^{6}}{6}.

$$\frac{x^{6}}{3}-12\int x^{3}\mathrm{d}x+18\int x\mathrm{d}x$$

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}.

$$\frac{x^{6}}{3}-3x^{4}+18\int x\mathrm{d}x$$

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

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

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.

$$9x^{2}-3x^{4}+\frac{x^{6}}{3}+С$$

Differentiate w.r.t. x

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