Integrate (x-2)/(x^2-4x+3)^3

How do I do this? Many thanks!

(x - 2) / (x^2 - 4x + 3)^3 dx

By u-substitution, let u = x^2 - 4x + 3. Then du = 2x - 2 dx, and 1/2 du = x - 2 dx. Now we can re-write this in terms of u to get:

1/2 * 1/u^3 du

By standard power-rule, the integral is:

-1/4 * 1/u^2 + C

Substituting back in for u, we have the final solution:

(-1 / 4(x^2 - 4x + 3)^2) + C

above ans is incorrect as d/dx((x^2 - 4x + 3)^3 = 2x - 4 and not 2x - 2 .

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u = (the term in the denominator)
du will end up being 2 multiplied by the numerator