Question

Solve

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

• A. $log\sqrt{x^2-4x+3+c}$
• B. $xlog(x-3)-2log(x-2)+c$
• C. $log\left[\right(x-3\left)\right(x-1\left)\right]$
• D. $Noneofthese$

Answer 1

Answer: (D)

$Noneofthese$

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

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

$=\int \frac{x-2}{(x-3)-1(x-3)}dx$

$=\int \frac{x-2}{(x-3)(x-1)}dx$

$=\int \frac{x-3+1}{(x-3)(x-1)}dx$

$=\int \frac{x-3}{(x-3)(x-1)}dx+\int \frac{1}{(x-3)(x-1)}dx$

$=\int \frac{dx}{x-1}+\frac{1}{2}\int [\frac{1}{(x-3)}-\frac{1}{x-1}]dx$

$=\log (x-1)+\frac{1}{2}\log \int \frac{1}{(x-3)}-\frac{1}{2}\log \int \frac{1}{(x-1)}$

$=\log (x-1)+\frac{1}{2}\log (x-3)-\frac{1}{2}\log (x-1)$

$=\frac{1}{2}\log (x-1)+\frac{1}{2}\log (x-3)+c$

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

Hence, the answer is $\log \sqrt{x^2-4x+3}+c.$