Question

Solve : $\int \frac{dx}{\sqrt{x^2-3x+2}}$

Answer 1

$\int \frac{dx}{\sqrt{x^2-3x+2}}=\int \frac{dx}{\sqrt{x^2-3x+\frac{9}{4}-\frac{9}{4}+2}}$

$=\int \frac{dx}{\sqrt{(x-\frac{3}{2}{)}^2-({\frac{1}{2}}^2\left)\right)}}$

$=log(x-\frac{3}{2}+\sqrt{(x-\frac{3}{2}{)}^2-\frac{1}{4}})+c$

$\int \because \frac{dx}{\sqrt{x^2-a^2}}=log(x+\sqrt{x^2-a^2})+c$

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