Posted by Jack on October 02, 2002 at 18:21:29:
In Reply to: help with this integral please posted by Joel on October 02, 2002 at 12:35:54:
: How do I approach this one?
: Integral of [x/(1+x^4)]dx
:Try this - Let w=x^2. Then dw=2xdx
:So I[x/(1+x^4)]dx = I [w/(1+w^2)]dw
:Now use integration by parts
:Let u=w or du=dw
:Let dv=dw/(1+w^2) or v=atanw
:(1/2)Iudv=(1/2)uv-(1/2)Ivdu = w*atanw-Iatanw dw
: =(1/2)w*atanw-(1/2)[w*atanw-(1/2)*ln(1+w^2)
: =(1/4)*ln(1+w^2)
:Sub in w=x^2 and get
: I[x/(1+x^4)]dx = (1/4)*ln(1+x^4)+C
NOTE: Using MATHCAD I get (1/2)*atan(x^2)