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:

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

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

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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

Let 2-x=t

differentiating both sides w.r.t.x

0 - 1 = \frac{dt}{dx}0−1=

dx

dt

dx = - dtdx=−dt

Integrating the function w.r.t.x

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

(2−x)

2

+t

1

dx

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

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

t

2

+1

−dt

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

t

2

+(1)

2

dt

It is the form of :

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

x

2

+a

2

1

dx=log∣x+

x

2

+a

2

∣+c

∴ Replace a by 1 and x by t we get

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

t

2

+1

∣+c

= $log { |t + \sqrt{ {t}^{2} + 1} | }^{ - 1} + c=log∣t+ </p><p>t$

2

+1

∣

−1

+c

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

∣t+

t

2

+1

∣

1

+c

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

∣2−x+

(2−x)

2

+1

∣

1

+c

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

∣2−x+

4+x

2

−4x+1

∣

1

+c

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

∣2−x+

x

2

−4x+5

∣

1

+c

╚════════════════════════╝

нσρє ıт нєłρs yσυ

_____________________

тнαηkyσυ