Please help me to evaluate the integral:
∫∞−1(x41+x6)2dx
Thanks.
Answer
Notice that:
x8=(x6+1)x2−x2
So that your integrand takes the form:
∫x21+x6dx−∫x2(1+x6)2dx
Now substitute u=x3,du=3x2dx:
∫13(1+u2)du−∫13(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)du−∫u23(1+u2)2du
I think you get the idea..
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