Solve ∫ left(3x^2-2right)^2 wrt x

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

Evaluate

$$\frac{9x^{5}}{5}-4x^{3}+4x+С$$

Solution Steps

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

$$\int 9\left(x^{2}\right)^{2}-12x^{2}+4\mathrm{d}x$$

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

$$\int 9x^{4}-12x^{2}+4\mathrm{d}x$$

Integrate the sum term by term.

$$\int 9x^{4}\mathrm{d}x+\int -12x^{2}\mathrm{d}x+\int 4\mathrm{d}x$$

Factor out the constant in each of the terms.

$$9\int x^{4}\mathrm{d}x-12\int x^{2}\mathrm{d}x+\int 4\mathrm{d}x$$

Since \int x^{k}\mathrm{d}x=\frac{x^{k+1}}{k+1} for k\neq -1, replace \int x^{4}\mathrm{d}x with \frac{x^{5}}{5}. Multiply 9 times \frac{x^{5}}{5}.

$$\frac{9x^{5}}{5}-12\int x^{2}\mathrm{d}x+\int 4\mathrm{d}x$$

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}.

$$\frac{9x^{5}}{5}-4x^{3}+\int 4\mathrm{d}x$$

Find the integral of 4 using the table of common integrals rule \int a\mathrm{d}x=ax.

$$\frac{9x^{5}}{5}-4x^{3}+4x$$

Simplify.

$$4x-4x^{3}+\frac{9x^{5}}{5}$$

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.

$$4x-4x^{3}+\frac{9x^{5}}{5}+С$$

Differentiate w.r.t. x

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