Question

integrate the function~​

Answer 1

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

Answer 2

Answer:

$\large \bold\red{ ln |4x - 6 + \sqrt{4 {x}^{2} - 12x + 5} | + C'}$

Step-by-step explanation:

Given,

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

Further simplifying,

We get,

$= \int \frac{1}{ \sqrt{ {x}^{2} - x - 2x + 2} } dx \\ \\ = \int \frac{1}{ \sqrt{ {x}^{2} - 3x + 2} } dx \\ \\ = \int\frac{1}{ \sqrt{ {(x)}^{2} - 2 \cdot \frac{3}{2} \cdot x + {( \frac{3}{2} )}^{2} + 2 - \frac{9}{4} } }dx \\ \\ = \int \frac{1}{ \sqrt{ {(x - \frac{3}{2}) }^{2} + ( \frac{8 - 9}{4} )} } dx \\ \\ = \int \frac{1}{ \sqrt{ {(x - \frac{3}{2}) }^{2} - {( \frac{1}{2} )}^{2} } } dx$

Now,

Let's assume that,

$x - \frac{3}{2} = t$

Differentiating both the sides,

We get,

$= > dx = dt$

Substituting the values,

We get,

$= \int \frac{1}{ \sqrt{ {(t)}^{2} - {( \frac{1}{2}) }^{2} } } dt \\ \\ = ln |t + \sqrt{ {t}^{2} - \frac{1}{4} } | + c\\ \\ = ln |t + \frac{ \sqrt{ {t}^{2} - 1 } }{2} | + c$

Substituting the values of t,

We get,

$= ln |x - \frac{3}{2} + \frac{ \sqrt{ {x}^{2} - 3x + \frac{9}{4} - 1 } }{2} | + c \\ \\ = ln |x - \frac{3}{2} + \frac{ \sqrt{4 {x}^{2} - 12x + 5} }{4} | + c \\ \\ ln | \frac{4x - 6 + \sqrt{4 {x}^{2} - 12x + 5} }{4} | + c \\ \\ = \large \bold{ ln |4x - 6 + \sqrt{4 {x}^{2} - 12x + 5} | + C'}$

Where,

C' is an integral Constant.