int frac{x - 2}{x^2 - 4x + 3} dx = | vedclass

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Question

(A) Let $I = \int \frac{x - 2}{x^2 - 4x + 3} dx$. Substitute $t = x^2 - 4x + 3$. Then $dt = (2x - 4) dx = 2(x - 2) dx$,which implies $(x - 2) dx = \fr

Vedclass · Accepted Answer

$\log \sqrt{x^2 - 4x + 3} + c$

Vedclass · Answer

(A) Let $I = \int \frac{x - 2}{x^2 - 4x + 3} dx$. Substitute $t = x^2 - 4x + 3$. Then $dt = (2x - 4) dx = 2(x - 2) dx$,which implies $(x - 2) dx = \frac{1}{2} dt$. Substituting these into the integral: $I = \int \frac{1}{2t} dt = \frac{1}{2} \log |t| + c$. Substituting back $t = x^2 - 4x + 3$: $I = \frac{1}{2} \log |x^2 - 4x + 3| + c = \log \sqrt{|x^2 - 4x + 3|} + c$. Thus,the correct option is $A$.

$\int \frac{x - 2}{x^2 - 4x + 3} dx = $

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