J jmf5x New member Joined Aug 28, 2006 Messages 1 Aug 28, 2006 #1 I have to integrate the following using integration by parts: . . .int [(ln(3x))^2] dx I just need help figuring how what to use! Please help!
I have to integrate the following using integration by parts: . . .int [(ln(3x))^2] dx I just need help figuring how what to use! Please help!
G galactus Super Moderator Staff member Joined Sep 28, 2005 Messages 7,203 Aug 28, 2006 #2 Let \(\displaystyle \L\\u=(ln(3x))^{2};\;\ dv=dx;\;\ du=\frac{2ln(3x)}{x}dx;\;\ v=x\) \(\displaystyle \L\\xln(3x)^{2}-\int{2ln(3x)}dx\) Use parts again: Let \(\displaystyle \L\\u=ln(3x);\;\ dv=dx;\;\ du=\frac{1}{x}dx;\;\ v=x\) \(\displaystyle \L\\xln(3x)^{2}-2\left[xln(3x)-x\int\frac{1}{x}dx\right]\) \(\displaystyle \L\\xln(3x)^{2}-2xln(3x)+2x\)
Let \(\displaystyle \L\\u=(ln(3x))^{2};\;\ dv=dx;\;\ du=\frac{2ln(3x)}{x}dx;\;\ v=x\) \(\displaystyle \L\\xln(3x)^{2}-\int{2ln(3x)}dx\) Use parts again: Let \(\displaystyle \L\\u=ln(3x);\;\ dv=dx;\;\ du=\frac{1}{x}dx;\;\ v=x\) \(\displaystyle \L\\xln(3x)^{2}-2\left[xln(3x)-x\int\frac{1}{x}dx\right]\) \(\displaystyle \L\\xln(3x)^{2}-2xln(3x)+2x\)
