How would you find the integration of ln(1+x) without integration by parts?? Thanks
J jeng New member Joined Apr 13, 2006 Messages 6 May 2, 2006 #1 How would you find the integration of ln(1+x) without integration by parts?? Thanks
pka Elite Member Joined Jan 29, 2005 Messages 11,976 May 2, 2006 #2 \(\displaystyle \L \int {\ln (u)du = u\ln (u) - u}\)
G galactus Super Moderator Staff member Joined Sep 28, 2005 Messages 7,216 May 2, 2006 #3 Integrate the power series for ln(1+x) \(\displaystyle \L\\\sum_{k=0}^{\infty}(-1)^{k}\frac{x^{k+1}}{k+1}= x-\frac{x^{2}}{2}+\frac{x^{3}}{3}-\frac{x^{4}}{4}+.............\)
Integrate the power series for ln(1+x) \(\displaystyle \L\\\sum_{k=0}^{\infty}(-1)^{k}\frac{x^{k+1}}{k+1}= x-\frac{x^{2}}{2}+\frac{x^{3}}{3}-\frac{x^{4}}{4}+.............\)
G galactus Super Moderator Staff member Joined Sep 28, 2005 Messages 7,216 May 2, 2006 #4 pka said: \(\displaystyle \L \int {\ln (u)du = u\ln (u) - u}\) Click to expand... I'm sorry, pka, but isn't that integration by parts?.
pka said: \(\displaystyle \L \int {\ln (u)du = u\ln (u) - u}\) Click to expand... I'm sorry, pka, but isn't that integration by parts?.
pka Elite Member Joined Jan 29, 2005 Messages 11,976 May 2, 2006 #5 Yes it is! But why do it any other way?
tkhunny Moderator Staff member Joined Apr 12, 2005 Messages 11,339 May 3, 2006 #7 pka said: \(\displaystyle \L \int {\ln (u)du = u\ln (u) - u}\) Click to expand... I love it. Do it by parts, or by any other available method, then simply memorize the result. After that, who cares where it came from?
pka said: \(\displaystyle \L \int {\ln (u)du = u\ln (u) - u}\) Click to expand... I love it. Do it by parts, or by any other available method, then simply memorize the result. After that, who cares where it came from?
