Friday, 19 December 2014

calculus - Solveintinfty1left(fracx41+x6right)2dx



Please help me to evaluate the integral:

1(x41+x6)2dx



Thanks.


Answer



Notice that:
x8=(x6+1)x2x2
So that your integrand takes the form:
x21+x6dxx2(1+x6)2dx
Now substitute u=x3,du=3x2dx:
13(1+u2)du13(1+u2)2du

The first integral is a multiple of arctan(u), and the second can be solved dividing again into two parts:
13(1+u2)2du=13(1+u2)duu23(1+u2)2du
I think you get the idea..


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