improper integral [x^2/(9+x^6)dx], x = -infinity to x = infinity.
x^2/(9+x^6) = x^2/(9+(x^2)^3) is an even function. Therefore, the given integral is equal to 2*integral (x^2/(9+x^3)) dx, x =0 to infinity.
Let us have a transformation x^3+9 = t. Then,
x^2dx = (1/3)dt. When x= 0, t =9 and x= infinity, t = infinity.
Therefore, the given integral = 2Integral {1/t}(1/3)dt t= 9 to t= inf.
=(2logt), t = 0 to t=inf.
= 2 log (infinity) - 2log9
= infinity.
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