Question

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

Answer 1

Solution:-

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

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

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

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

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

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

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

Answer 2

Answer:

Solution:-

Let 2-x=t

differentiating both sides w.r.t.x

0 - 1 = \frac{dt}{dx}0−1=dxdt

dx = - dtdx=−dt

Integrating the function w.r.t.x

∫ \frac{1}{ \sqrt{ {(2 - x)}^{2} + t} }∫(2−x)2+t1

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

= ∫ \frac{ - dt}{ \sqrt{ {t}^{2} + 1} }=∫t2+1−dt

= - ∫ \frac{dt}{ \sqrt{ {t}^{2} + {(1)}^{2} } }=−∫t2+(1)2dt

It is the form of :

∫ \frac{1}{ \sqrt{ {x}^{2} + {a}^{2} } } dx = log |x + \sqrt{ {x}^{2} + {a}^{2} } |∫x2+a21dx=log∣x+x2+a2∣

∴ Replace a by 1 and x by t we get

= - log |t + \sqrt{ {t}^{2} + 1 } |=−log∣t+t2+1∣

= log { |t + \sqrt{ {t}^{2} + 1} | }^{ - 1}=log∣t+t2+1∣−1

= log \frac{1}{ |t + \sqrt{ {t}^{2} + 1} | }=log∣t+t2+1∣1

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

= log \frac{1}{ |2 - x + \sqrt{4 + {x}^{2} - 4x + 1} | }=log∣2−x+4+x2−4x+1∣1

= log \frac{1}{ |2 - x + \sqrt{ {x}^{2} - 4x + 5} | }=log∣2−x+x2−4x+5∣1