C cmnalo Junior Member Joined Nov 5, 2006 Messages 61 Dec 3, 2006 #1 ∫x+2 / x^2 +4x +3 dx u= x^2 + 4x +3 du= 2x +4dx dx= 1 /2x +4dx I'm not sure if I'm approaching this problem correctly? I'm also not sure about what the next step would be. ∫? / u Answer: (1/2)ln│x^2 +4x+3│+c
∫x+2 / x^2 +4x +3 dx u= x^2 + 4x +3 du= 2x +4dx dx= 1 /2x +4dx I'm not sure if I'm approaching this problem correctly? I'm also not sure about what the next step would be. ∫? / u Answer: (1/2)ln│x^2 +4x+3│+c
A arthur ohlsten Full Member Joined Feb 20, 2005 Messages 847 Dec 3, 2006 #2 S[x+2]dx /[x^2+4x+3] let u=x^2+4x+3 then du=[2x+4]dx du=2[x+2] dx or dx= du/[2(x+2)] substitute S [x+2] du/{2(x+2) u} S du/[2u] 1/2 S du/u 1/2 lnu +c 1/2 ln[x^2+4x+3] +c answer Arthur
S[x+2]dx /[x^2+4x+3] let u=x^2+4x+3 then du=[2x+4]dx du=2[x+2] dx or dx= du/[2(x+2)] substitute S [x+2] du/{2(x+2) u} S du/[2u] 1/2 S du/u 1/2 lnu +c 1/2 ln[x^2+4x+3] +c answer Arthur
