Question

Please answer this question​

Answer 1

-ln(| x - 1 |) + -ln(| x - 2 |) + C

Explanation:

Answer 2

`\displaystyle \int (1)/(x^2 -3x +2) dx\\\\\\=\displaystyle \int (dx)/(x^2 -3\cdot \frac 32 \cdot x + \left(\frac 32 \right)^2 -\left( \frac 32 \right)^2 +2)\\\\\\=\displaystyle \int (dx)/(\left(x- \frac 32 \right)^2 +2- \frac 94)\\\\\\=\displaystyle \int (dx)/(\left(x - \frac 32 \right)^2-\frac 14 )\\\\\\=\displaystyle \int (du)/(u^2-\left( \frac 12 \right)^2) ~~~~~~~~~~~~~~~~~~;\left[u = x - \frac32\implies du= dx\right]\\`

`=(1)/(2\cdot\tfrac 12) \ln \left| (u-\frac 12 )/(u+\frac 12 ) \right|+C~~~~~~~~~~:\left[ \displaystyle \int \frac {dx}{x^2 -a^2} = \frac 1{2a} \ln \left| (x-a)/(x+a) \right| +C\right]\\\\\\=\ln\left|(x- \frac 32 - \frac 12)/(x-\frac3 2 + \frac 12 ) \right|+C~~~~~~~~~~~~;\left[\text{Substitute back}~u =x- \frac 32 \right]\\\\\\=\ln\left|(x-2)/(x-1) \right|+C`