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January 25, 2015

January 25, 2015

Posted by **mathstudent** on Wednesday, November 7, 2007 at 4:40pm.

- math -
**drwls**, Thursday, November 8, 2007 at 8:46pmThere 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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