Solve ∫ (3-x^2)^2 wrt x

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

Evaluate

$$\frac{x^{5}}{5}-2x^{3}+9x+С$$

Solution Steps

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

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

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

$$\int 9-6x^{2}+x^{4}\mathrm{d}x$$

Integrate the sum term by term.

$$\int 9\mathrm{d}x+\int -6x^{2}\mathrm{d}x+\int x^{4}\mathrm{d}x$$

Factor out the constant in each of the terms.

$$\int 9\mathrm{d}x-6\int x^{2}\mathrm{d}x+\int x^{4}\mathrm{d}x$$

Find the integral of 9 using the table of common integrals rule \int a\mathrm{d}x=ax.

$$9x-6\int x^{2}\mathrm{d}x+\int x^{4}\mathrm{d}x$$

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 -6 times \frac{x^{3}}{3}.

$$9x-2x^{3}+\int x^{4}\mathrm{d}x$$

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}.

$$9x-2x^{3}+\frac{x^{5}}{5}$$

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.

$$\frac{x^{5}}{5}-2x^{3}+9x+С$$

Differentiate w.r.t. x

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