Solve ∫ (4-x^2)^2x^2 wrt x=

$$\int ( 4 - x ^ { 2 } ) ^ { 2 } x ^ { 2 } d x =$$

Evaluate

$$\frac{x^{7}}{7}-\frac{8x^{5}}{5}+\frac{16x^{3}}{3}+С$$

Solution Steps

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(4-x^{2}\right)^{2}.

$$\int \left(16-8x^{2}+\left(x^{2}\right)^{2}\right)x^{2}\mathrm{d}x$$

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

$$\int \left(16-8x^{2}+x^{4}\right)x^{2}\mathrm{d}x$$

Use the distributive property to multiply 16-8x^{2}+x^{4} by x^{2}.

$$\int 16x^{2}-8x^{4}+x^{6}\mathrm{d}x$$

Integrate the sum term by term.

$$\int 16x^{2}\mathrm{d}x+\int -8x^{4}\mathrm{d}x+\int x^{6}\mathrm{d}x$$

Factor out the constant in each of the terms.

$$16\int x^{2}\mathrm{d}x-8\int x^{4}\mathrm{d}x+\int x^{6}\mathrm{d}x$$

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

$$\frac{16x^{3}}{3}-8\int x^{4}\mathrm{d}x+\int x^{6}\mathrm{d}x$$

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 -8 times \frac{x^{5}}{5}.

$$\frac{16x^{3}}{3}-\frac{8x^{5}}{5}+\int x^{6}\mathrm{d}x$$

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}.

$$\frac{16x^{3}}{3}-\frac{8x^{5}}{5}+\frac{x^{7}}{7}$$

Simplify.

$$\frac{x^{7}}{7}-\frac{8x^{5}}{5}+\frac{16x^{3}}{3}$$

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.

$$\frac{x^{7}}{7}-\frac{8x^{5}}{5}+\frac{16x^{3}}{3}+С$$

Differentiate w.r.t. x

$$\left(x\left(x^{2}-4\right)\right)^{2}$$