Integration by substitution

[cmnalo]

∫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

 

[arthur ohlsten]

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