Mathematics

Evaluate $$\int \dfrac{{dx}}{{\sqrt {{{\left( {2 - x} \right)}^2} - 1} }}$$


SOLUTION

Consider the given integral.

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

 

Let $$t=2-x$$

$$ dt=-dx $$

$$ dx=-dt $$

 

Therefore,

$$I=-\int{\dfrac{dt}{\sqrt{{{t}^{2}}-1}}}$$

 

We know that,

$$\int{\dfrac{dx}{\sqrt{{{x}^{2}}-{{a}^{2}}}}}=\dfrac{1}{2a}\log \left| \dfrac{x-a}{x+a} \right|+c$$

 

Therefore,

$$I=-\dfrac{1}{2}\log \left| \dfrac{t-1}{t+1} \right|+c$$

 

On putting the value of t, we get

$$ I=-\dfrac{1}{2}\log \left| \dfrac{2-x-1}{2-x+1} \right|+c $$

$$ I=-\dfrac{1}{2}\log \left| \dfrac{1-x}{3-x} \right|+c $$

$$ I=\dfrac{1}{2}\log \left| \dfrac{3-x}{1-x} \right|+c $$

 

Hence, the value of integral is $$\dfrac{1}{2}\log \left| \dfrac{3-x}{1-x} \right|+c$$.

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