Question

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

Answer 1

Step-by-step explanation:

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

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

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

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

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_ _ _ _ _ _ _ _ _ _ _ _ _ _ _✍️

$\bold{ Let \:2-x=t}$

$\bold{differentiating \:both\: sides \:w.r.t.x}$

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

$dx = - dt$

Integrating the function w.r.t.x

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

$\bold{\red{Put\: the\: value \:of \:2-x=t \:and \:dx=-dt}$

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

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

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

Step-by-step explanation:

☘

℘ɧεŋσɱεŋศɭ

☘

\red{\bold{\underline{\underline{❥Question᎓}}}}

❥Question᎓

integrate the function:

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

(2−x)

2

+1

1

\huge\tt\underline\blue{❯Answer❮ }

❯Answer❮

╔════════════════════════╗

_ _ _ _ _ _ _ _ _ _ _ _ _ _ _✍️

\bold{ Let \:2-x=t}Let2−x=t

\bold{differentiating \:both\: sides \:w.r.t.x}differentiatingbothsidesw.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

\bold{\red{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+

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