B blue27 New member Joined Feb 14, 2018 Messages 2 Feb 14, 2018 #1 Use integration by parts to evaluate this integral: . . . . .\(\displaystyle \displaystyle \int_0^3\, \ln(x^2\, +\, 9)\, dx\) Attachments Screenshot_6.png 2.3 KB · Views: 9 Last edited by a moderator: Feb 16, 2018
Use integration by parts to evaluate this integral: . . . . .\(\displaystyle \displaystyle \int_0^3\, \ln(x^2\, +\, 9)\, dx\)
tkhunny Moderator Staff member Joined Apr 12, 2005 Messages 11,325 Feb 15, 2018 #2 blue27 said: Use integration by parts to evaluate this integral: . . . . .\(\displaystyle \displaystyle \int_0^3\, \ln(x^2\, +\, 9)\, dx\) Click to expand... Okay, what have you tried? Perhaps the basic formulation? Here's the Indefinite Version \(\displaystyle \int\ln\left(x^{2}+9\right)\;dx = x\cdot\ln\left(x^{2}+9\right)-\int\;x\;d\left[\ln\left(x^{2}+9\right)\right]\) Where do we go from there? Last edited by a moderator: Feb 16, 2018
blue27 said: Use integration by parts to evaluate this integral: . . . . .\(\displaystyle \displaystyle \int_0^3\, \ln(x^2\, +\, 9)\, dx\) Click to expand... Okay, what have you tried? Perhaps the basic formulation? Here's the Indefinite Version \(\displaystyle \int\ln\left(x^{2}+9\right)\;dx = x\cdot\ln\left(x^{2}+9\right)-\int\;x\;d\left[\ln\left(x^{2}+9\right)\right]\) Where do we go from there?