Posted by mathstudent on Wednesday, November 7, 2007 at 4:40pm.
There is a reduction formula for this. The number of terms in the answer depends upon the number k.
The integral is
{1/[2(2k+1)]}*[2x*(x^2+1)^k + 4k S (x^2+1)^(k-1) dx]
Where S represents an integral sign.
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