May 4, 2010 #2 jakncoke Messages 16 Reaction score 0 Applying integration by parts which states \int u\frac{dv}{dx}\,dx = uv - \int v \frac{du}{dx}\,dx u=ln(x) and v=x So \int ln(x)\,dx = xln(x) - \int x\frac{1}{x}\,dx = xln(x)-x+c
Applying integration by parts which states \int u\frac{dv}{dx}\,dx = uv - \int v \frac{du}{dx}\,dx u=ln(x) and v=x So \int ln(x)\,dx = xln(x) - \int x\frac{1}{x}\,dx = xln(x)-x+c
May 4, 2010 #3 bmed90 Messages 99 Reaction score 0 I wouldn't have thought to use parts. Thanks, I appreciate it