- #1
Ocis
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Determine the integral (x^2) / (2x^3 - 3)^2
This is probably easier than I expect but I have been attempting it using substitution...?
u = (2x^3 -3)
du/dx = 6x^2
dx = du / 6x^2
x^2 / (u^2) . du / 6x^2 - then I lose it because of the canceling out, and I'm unsure what happens to the x's...
I know the answer is -1/6 (2x^3 - 3) + C, but would appreciate the working I miss out. Or if I am attempting it using the wrong methods.
Many thanks,
Ocis
This is probably easier than I expect but I have been attempting it using substitution...?
u = (2x^3 -3)
du/dx = 6x^2
dx = du / 6x^2
x^2 / (u^2) . du / 6x^2 - then I lose it because of the canceling out, and I'm unsure what happens to the x's...
I know the answer is -1/6 (2x^3 - 3) + C, but would appreciate the working I miss out. Or if I am attempting it using the wrong methods.
Many thanks,
Ocis