Tìm ∫1/x2−4x+3dx

Question

Tìm \(\int \frac{1}{x^{2} -4x+3} {\rm d}x \)

 A.\( \frac{1}{2} \ln \left|\frac{x-1}{x-3} \right|+C. \)

B.\(\frac{1}{2} \ln \left|\frac{x-3}{x-1} \right|+C. \)

C. \(\ln \left|x^{2} -4x+3\right|+C. \)

D. \(\ln \left|\frac{x-3}{x-1} \right|+C.\)

Answer 1

Chọn B

 Ta có: \(\int \frac{1}{x^{2} -4x+3} {\rm d}x =\int \frac{1}{(x-1)(x-3)} {\rm d}x =\frac{1}{2} \left(\int \frac{1}{x-3} {\rm d}x -\int \frac{1}{x-1} {\rm d}x \right)\)
\(=\frac{1}{2} \left(\ln \left|x-3\right|-\ln \left|x-1\right|\right)=\frac{1}{2} \ln \left|\frac{x-3}{x-1} \right|+C.\)