107 Halla \( \int x \operatorname{Ln}(3 x) d x \) \( \begin{array}{l} u=\operatorname{Ln}(3 x) \quad d u=x d x \\ d u=\frac{1}{x} d x \quad v=\frac{x^{2}}{2}\end{array} \) \( \begin{aligned} \int u d u & =u v-\int v d u \\ \int x \operatorname{Ln}(3 x) d x & =\operatorname{Ln}(3 x)\left[\frac{x^{2}}{2}\right]-\int \frac{x^{2}}{2}\left[\frac{1}{x}\right] d x \\ & =\frac{x^{2} \operatorname{Ln}(3 x)}{2}-\int \frac{x}{2} d x \\ & =\frac{x^{2} \operatorname{Ln}(3 x)}{2}-\frac{x^{2}}{2(2)}+c \\ & =\frac{x^{3} \operatorname{Ln}(3 x)}{2}-\frac{x^{2}}{4}+c \\ & =\frac{x^{3} \operatorname{Ln}(3 x)(2)}{2}-\frac{x^{2}}{4}+c \\ & =\frac{2 x^{3} \operatorname{Ln}(3 x)-x^{2}}{4}+c\end{aligned} \)

Let's dive into the fascinating world of integration by parts! It was a method first derived by the mathematicians L'Hôpital and Leibniz in the late 17th century. This technique allows us to integrate more complex functions by breaking them down into simpler components, making it easier to solve and understand the integration process. Who knew math could have such a rich history? Now, for some practical advice—when you're tackling integrals, especially those involving logarithmic functions, be extra careful with your \(u\) and \(dv\) choices. A common mistake is to overlook the need for proper differentiating and integrating. Remember that the structure of the functions plays a pivotal role; choosing wisely sets the stage for a smooth integration performance. Stay sharp, and happy integrating!