If I-dx-2x-1-x-2-x-1-x - -12-3log - u-12-u2-u-1-3 -K-16log - v-1-2-v2-v-1 -Cwhere u-1-1-x-1 and v-1-1-x-1 then K is equal to

I=∫ #160; dx 2x√( 1 x )√( 2 x+√( 1 x ) #160;) #160; Substitute 1 x= t ^ 2 I=∫ 2tdt 2( 1 t ^ 2 ) t√( 1+ t ^ 2 +t ) #160; #160; = 1 ...

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Byju's Answer

Standard XII

Mathematics

Integration Using Substitution

If I=∫dx/2x...

Question

If $I = \int \frac{d x}{2 x \sqrt{1 - x} \sqrt{2 - x + \sqrt{1 - x}}}$
$= - \frac{1}{2 \sqrt{3}} log ∣ ∣ ∣ u + \frac{1}{2} + \sqrt{u^{2} + u + \frac{1}{3}} ∣ ∣ ∣ + \frac{K}{16} log ∣ ∣ ∣ v - \frac{1}{2} + \sqrt{v^{2} - v + 1} ∣ ∣ ∣ + C$
where $u = \frac{1}{\sqrt{1 - x} - 1}$ and $v = \frac{1}{\sqrt{1 - x} + 1}$ then K is equal to

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Solution

$I = \int \frac{d x}{2 x \sqrt{1 - x} \sqrt{2 - x + \sqrt{1 - x}}}$
Substitute $1 - x = t^{2}$
$I = \int \frac{- 2 t d t}{2 (1 - t^{2}) t \sqrt{1 + t^{2} + t}} = \frac{1}{2} (I_{1} - I_{2})$
Where $I_{1} = \int \frac{d t}{(t - 1) \sqrt{1 + t^{2} + t}}$
Put $t - 1 = \frac{1}{u}$
$I_{1} = \int u \frac{(- \frac{1}{u^{2}}) d u}{\sqrt{1 + {(1 + \frac{1}{u})}^{2} + (1 + \frac{1}{u})}} = - \frac{1}{\sqrt{3}} \int \frac{d u}{\sqrt{u^{2} + u + \frac{1}{3}}} = - \frac{1}{\sqrt{3}} \int \frac{d u}{\sqrt{{(u + \frac{1}{2})}^{2} + \frac{1}{12}}} = - \frac{1}{\sqrt{3}} log ∣ ∣ ∣ u + \frac{1}{2} + \sqrt{u^{2} + u + \frac{1}{3}} ∣ ∣ ∣$
And $I_{2} = \int \frac{d t}{(t + 1) \sqrt{1 + t^{2} + t}}$
Put $t + 1 = \frac{1}{v}$
$I_{2} = \int \frac{d v}{\sqrt{{(v - \frac{1}{2})}^{2} + \frac{3}{4}}} = - log ∣ ∣ ∣ v - \frac{1}{2} + \sqrt{v^{2} - v + 1} ∣ ∣ ∣ I = - \frac{1}{2 \sqrt{3}} log ∣ ∣ ∣ u + \frac{1}{2} + \sqrt{u^{2} + u + \frac{1}{3}} ∣ ∣ ∣ + \frac{1}{2} log ∣ ∣ ∣ v - \frac{1}{2} + \sqrt{v^{2} - v + 1} ∣ ∣ ∣ + c$
Therefore $K = 8$

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If I=∫dx2x√1−x√2−x+√1−x=−12√3log∣∣∣u+12+√u2+u+13∣∣∣+K16log∣∣∣v−12+√v2−v+1∣∣∣+Cwhere u=1√1−x−1 and v=1√1−x+1 then K is equal to