Question

integrate the function:
$\frac{1}{ \sqrt{ {(2 - x)}^{2} + 1} }$

Answer 1

Step-by-step explanation:

Step-by-step explanation:

$\huge\mathfrak\green{\bold{\underline{☘{ ℘ɧεŋσɱεŋศɭ}☘}}}$

$\red{\bold{\underline{\underline{❥Question᎓}}}}$integrate the function:

$\frac{1}{ \sqrt{ {(2 - x)}^{2} + 1} }$

$\huge\tt\underline\blue{❯Answer❮</p><p> }$

╔════════════════════════╗

_ _ _ _ _ _ _ _ _ _ _ _ _ _ _✍️

Let 2-x=t

differentiating both sides w.r.t.x

$0 - 1 = \frac{dt}{dx}$

$dx = - dt$

Integrating the function w.r.t.x

$∫ \frac{1}{ \sqrt{ {(2 - x)}^{2} + t} } dx$

Put the value of 2-x=t and dx=-dt

$= ∫ \frac{ - dt}{ \sqrt{ {t}^{2} + 1} }$

$= - ∫ \frac{dt}{ \sqrt{ {t}^{2} + {(1)}^{2} } }$

It is the form of :

$∫ \frac{1}{ \sqrt{ {x}^{2} + {a}^{2} } } dx = log |x + \sqrt{ {x}^{2} + {a}^{2} } | + c$

∴ Replace a by 1 and x by t we get

$= - log |t + \sqrt{ {t}^{2} + 1 } | + c$

$= log { |t + \sqrt{ {t}^{2} + 1} | }^{ - 1} + c$

$= log \frac{1}{ |t + \sqrt{ {t}^{2} + 1} | } + c$

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

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

$= log \frac{1}{ |2 - x + \sqrt{ {x}^{2} - 4x + 5} | } + c$

╚════════════════════════╝

нσρє ıт нєłρs yσυ

_____________________

тнαηkyσυ

Answer 2

Step-by-step explanation:

$\sf= - log |t + \sqrt{ {t}^{2} + 1 } | + c$

$\sf= log { |t + \sqrt{ {t}^{2} + 1} | }^{ - 1} + c$

$\sf \red{= log \frac{1}{ |2 - x + \sqrt{ {x}^{2} - 4x + 5} | } + c}$