Question

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

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

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

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

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

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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

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

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

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

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

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

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

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Answer 2

$\int \frac{1}{\sqrt{\left(2-x\right)^2+1}}dx=\ln \left|\sqrt{x^2-4x+5}+x-2\right|+C$

$\mathrm{Expand}\:\left(2-x\right)^2+1:\quad x^2-4x+5$

$=\int \frac{1}{\sqrt{x^2-4x+5}}dx$

$\mathrm{Complete\:the\:square}\:x^2-4x+5:\quad \left(x-2\right)^2+1$

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

$\mathrm{Apply\:u-substitution:}\:u=x-2$

$=\int \frac{1}{\sqrt{u^2+1}}du$

$\mathrm{Use\:the\:common\:integral}:\quad \int \frac{1}{\sqrt{u^2+1}}du=\ln \left|\sqrt{u^2+1}+u\right|$

$=\ln \left|\sqrt{u^2+1}+u\right|$

$\mathrm{Substitute\:back}\:u=x-2$

$=\ln \left|\sqrt{\left(x-2\right)^2+1}+x-2\right|$

$\mathrm{Expand}\:\left(x-2\right)^2+1:\quad x^2-4x+5$

$=\ln \left|\sqrt{x^2-4x+5}+x-2\right|$

$\mathrm{Add\:a\:constant\:to\:the\:solution}$

$=\ln \left|\sqrt{x^2-4x+5}+x-2\right|+C$