Integration F.6

∫x/(1+x)dx
=(1/2)∫1/(1+x)d(1+x)
=(1/2)ln|1+x|+C, where C is a constant

2008-07-02 23:22:48 補充：
yahoo出唔到某D字元，我打多次
∫x/(1+x²)dx

=(1/2)∫1/(1+x²)d(1+x²)

=(1/2)ln|1+x²|+C, where C is a constant

Another substitution method:
http://i117.photobucket.com/albums/o61/billy_hywung/Jul08/Crazyint1.jpg

let x=tanθ

∫ x / ( 1+x^2 ) dx
=∫ tanθ / ( sec^2θ ) (sec^2θ) dθ
=∫ tanθ dθ
=∫-1/cosθ d (cosθ)
=-ln(cosθ)+C
=-ln(1/√(1+x^2))+C
=ln(√(1+x^2))+C