Solve ∫ (x^2-2)(x^2+2) wrt x

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

Evaluate

$$\frac{x^{5}}{5}-4x+С$$

Solution Steps

Consider \left(x^{2}-2\right)\left(x^{2}+2\right). Multiplication can be transformed into difference of squares using the rule: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}. Square 2.

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

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

$$\int x^{4}-4\mathrm{d}x$$

Integrate the sum term by term.

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

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}.

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

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

$$\frac{x^{5}}{5}-4x$$

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.

$$\frac{x^{5}}{5}-4x+С$$

Differentiate w.r.t. x

$$x^{4}-4$$