Consider $\left(x^{2}-y^{2}\right)\left(x^{2}+y^{2}\right)$. Multiplication can be transformed into difference of squares using the rule: $\left(a-b\right)\left(a+b\right)=a^{2}-b^{2}$.
To raise a power to another power, multiply the exponents. Multiply $2$ and $2$ to get $4$.
$$\int x^{4}-\left(y^{2}\right)^{2}\mathrm{d}x$$
To raise a power to another power, multiply the exponents. Multiply $2$ and $2$ to get $4$.
$$\int x^{4}-y^{4}\mathrm{d}x$$
Integrate the sum term by term.
$$\int x^{4}\mathrm{d}x+\int -y^{4}\mathrm{d}x$$
Factor out the constant in each of the terms.
$$\int x^{4}\mathrm{d}x-\int y^{4}\mathrm{d}x$$
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 $\frac{x^{5}}{5}$.
$$\frac{x^{5}}{5}-\int y^{4}\mathrm{d}x$$
Find the integral of $y^{4}$ using the table of common integrals rule $\int a\mathrm{d}x=ax$.
$$\frac{x^{5}}{5}-y^{4}x$$
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.
