Integration using u-substitution \[ \begin{array}{l|l} \text { 8. } \int\left(2 x^{5}\right)\left(x^{2}-3\right)^{5} d x & \text { Let } u=x \\ =\int\left(2 x^{5}\right)(u)^{5} \frac{d u}{2 x} & \frac{d u}{d x}=2 x \\ =\int x^{4}(u)^{5} d u=\int\left(x^{2}\right)^{2}(u)^{5} d u \\ =\int(u+3)^{2} u^{5} d u \quad=\int\left(u^{2}+6 u+9\right) u^{5} d u \\ =\int\left(u^{7}+6 u^{6}+9 u^{5}\right) d u & d x \\ =\frac{u^{8}}{8}+6 \frac{u^{7}}{7}+9 \frac{u^{6}}{6}+C \quad=\frac{\left(x^{2}+3\right)^{8}}{8}+6 \frac{\left(x^{2}+3\right)^{7}}{7}+9 \frac{\left(x^{2}+3\right)^{6}}{6}+C \end{array} \]

Did you know that "u-substitution" is like giving your integral a new name? By changing the variable, you often simplify complex integrals into more manageable ones. It's as if you're putting on a fresh pair of glasses to make everything clearer! Just remember that the trick is to pick a "u" that will simplify the function, and don't forget to calculate the derivative to change `dx` accordingly! When tackling integrals, one common mistake is forgetting to adjust the limits of integration if you're working with definite integrals. If you substitute `u`, you need to update the limits based on the original variable's limits too! This little oversight can lead to completely different answers, which is definitely not what you want. Always double-check your work, and you'll be on your way to mastering u-substitution!