Solve ∫ x^4(4x^2-3)^2

$$\int x ^ { 4 } ( 4 x ^ { 2 } - 3 ) ^ { 2 }$$

Evaluate

$$\frac{16x^{9}}{9}-\frac{24x^{7}}{7}+\frac{9x^{5}}{5}+С$$

Solution Steps

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(4x^{2}-3\right)^{2}.

$$\int x^{4}\left(16\left(x^{2}\right)^{2}-24x^{2}+9\right)\mathrm{d}x$$

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

$$\int x^{4}\left(16x^{4}-24x^{2}+9\right)\mathrm{d}x$$

Use the distributive property to multiply x^{4} by 16x^{4}-24x^{2}+9.

$$\int 16x^{8}-24x^{6}+9x^{4}\mathrm{d}x$$

Integrate the sum term by term.

$$\int 16x^{8}\mathrm{d}x+\int -24x^{6}\mathrm{d}x+\int 9x^{4}\mathrm{d}x$$

Factor out the constant in each of the terms.

$$16\int x^{8}\mathrm{d}x-24\int x^{6}\mathrm{d}x+9\int x^{4}\mathrm{d}x$$

Since \int x^{k}\mathrm{d}x=\frac{x^{k+1}}{k+1} for k\neq -1, replace \int x^{8}\mathrm{d}x with \frac{x^{9}}{9}. Multiply 16 times \frac{x^{9}}{9}.

$$\frac{16x^{9}}{9}-24\int x^{6}\mathrm{d}x+9\int x^{4}\mathrm{d}x$$

Since \int x^{k}\mathrm{d}x=\frac{x^{k+1}}{k+1} for k\neq -1, replace \int x^{6}\mathrm{d}x with \frac{x^{7}}{7}. Multiply -24 times \frac{x^{7}}{7}.

$$\frac{16x^{9}}{9}-\frac{24x^{7}}{7}+9\int x^{4}\mathrm{d}x$$

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}. Multiply 9 times \frac{x^{5}}{5}.

$$\frac{16x^{9}}{9}-\frac{24x^{7}}{7}+\frac{9x^{5}}{5}$$

Simplify.

$$\frac{9x^{5}}{5}-\frac{24x^{7}}{7}+\frac{16x^{9}}{9}$$

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.

$$\frac{9x^{5}}{5}-\frac{24x^{7}}{7}+\frac{16x^{9}}{9}+С$$

Differentiate w.r.t. x

$$x^{4}\left(4x^{2}-3\right)^{2}$$