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evaluate the integral of 2x^2-4x/x^3-2x^2 dx
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Actually might be simpler than you realize. Simply factor the numerator and denominator. 2x^2 - 4x factors into 2x(x-2) while the denominator factors to x^2(x-2). Therefore you'll see the (x-2) factors will cancel leaving you with integral 2x/x^2 dx. From there it should be pretty simple?
did you try using partial fractions?
nevermind, i totally missed the comment by Psymon. He explained it pretty well.
its ok but thank you anyways
did you figure it out?
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\[\frac{2 x^2-4 x}{x^3-2 x^2}=\frac{2}{x} \]
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