The value of int frac{dx}{sqrt{1 - x}} is | vedclass

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Question

(B) Let $I = \int \frac{dx}{\sqrt{1 - x}}$.We can rewrite the integral as $I = \int (1 - x)^{-1/2} dx$.Using the integration formula $\int (ax + b)^n

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$-2\sqrt{1 - x} + c$

Vedclass · Answer

(B) Let $I = \int \frac{dx}{\sqrt{1 - x}}$.We can rewrite the integral as $I = \int (1 - x)^{-1/2} dx$.Using the integration formula $\int (ax + b)^n dx = \frac{(ax + b)^{n+1}}{a(n+1)} + c$,where $a = -1$,$b = 1$,and $n = -1/2$:$I = \frac{(1 - x)^{-1/2 + 1}}{(-1)(-1/2 + 1)} + c$$I = \frac{(1 - x)^{1/2}}{(-1)(1/2)} + c$$I = \frac{\sqrt{1 - x}}{-1/2} + c$$I = -2\sqrt{1 - x} + c$.

The value of $\int \frac{dx}{\sqrt{1 - x}}$ is

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