Math, asked by mohammad7754, 6 months ago

integrate the function:
\frac{1}{ \sqrt{ {(2 - x)}^{2} + 1} }\  \textless \ br /\  \textgreater \

Answers

Answered by Anonymous
44

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} }

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

= ∫ \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} } |

∴ Replace a by 1 and x by t we get

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

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

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

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

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

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


Anonymous: Perfect ^^"
Answered by Anonymous
0

-by-step

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

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

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

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

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

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