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Q.Find ∫dx1+x−x\int \frac{dx}{\sqrt{1+x-\sqrt{x}}}.

Rajasthan RbseRajasthan Board Senior Secondary Examination 2018Subjective· 2mImportance★★★★★
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Substitute t=xt=\sqrt x to turn the integrand into a quadratic under the square root, then split into a perfect-derivative part and a standard ∫dtt2+shift\int\frac{dt}{\sqrt{t^2+\text{shift}}} part.

Let t=xt=\sqrt x, so x=t2x=t^2 and dx=2t dtdx=2t\,dt.

1+x−x=1+t2−t=t2−t+11+x-\sqrt x = 1+t^2-t = t^2-t+1

The integral becomes:

∫2t dtt2−t+1\displaystyle\int \dfrac{2t\,dt}{\sqrt{t^2-t+1}}

Write 2t=(2t−1)+12t = (2t-1)+1:

=∫2t−1t2−t+1 dt+∫1t2−t+1 dt= \displaystyle\int \dfrac{2t-1}{\sqrt{t^2-t+1}}\,dt + \int \dfrac{1}{\sqrt{t^2-t+1}}\,dt

For the first part, since ddt(t2−t+1)=2t−1\frac{d}{dt}(t^2-t+1)=2t-1:

∫2t−1t2−t+1 dt=2t2−t+1\displaystyle\int \dfrac{2t-1}{\sqrt{t^2-t+1}}\,dt = 2\sqrt{t^2-t+1}

For the second part, complete the square: t2−t+1=(t−12)2+34t^2-t+1 = \left(t-\tfrac12\right)^2+\tfrac34, and use the standard result ∫duu2+a2=ln⁡∣u+u2+a2∣+C\int\frac{du}{\sqrt{u^2+a^2}}=\ln|u+\sqrt{u^2+a^2}|+C:

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