No good...


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Posted by Joel on October 02, 2002 at 21:42:13:

In Reply to: Re: help with this integral please posted by Jack on October 02, 2002 at 18:21:29:

: : 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

small mistake there; should be: I[x/(1+x^4)]dx = (1/2)*I [1/(1+w^2)]dw

Now let Q = arctan(w)
so w = tan(Q)
and dw = sec^2(Q)dQ
and 1 + w^2 = 1 + tan^2(Q) = sec^2(Q)
now (1/2)*I [1/(1+w^2)]dw = (1/2) I [(sec^2(Q)dQ)/sec^2(Q)] = (1/2) I dQ
= (1/2) Q = (1/2) arctan(u) = (1/2) arctan (x^2) + C

(which is what you got with mathcad)


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