Evaluate $\int \dfrac{1}{\sqrt{1-x}}\,dx$

Question

Find $$\int \dfrac{1}{\sqrt{1-x}}\,dx$$

I did this and got $\dfrac23(1-x)^{\frac32} + c$

But a online calculator is telling me it should be $-2(1-x)^{\frac12}$

What one is on the money and if not me why?

Answer 1

$$ \int \frac{1}{\sqrt{1-x}}~dx=\int(1-x)^{-1/2}~dx $$ Let $u=1-x$, $du=-dx$, so $$ \int(1-x)^{-1/2}~dx=-\int u^{-1/2}~du $$ Add one to the power of $u$, and divide by the new power $$ \int \frac{1}{\sqrt{1-x}}~dx=-\int u^{-1/2}~du=-\frac{u^{1/2}}{1/2}+c=-2(1-x)^{1/2}+c $$

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What you have done is integrated $\sqrt{1-x}$ by mistake $$ \int \sqrt{1-x}~dx=\int (1-x)^{1/2}~dx=-\int u^{1/2}~du=-\frac{2}{3}u^{3/2}+c=-\frac{2}{3}(1-x)^{3/2}+c $$

Answer 2

$$\int\frac1{\sqrt x}dx=2\sqrt x+C\implies \int\frac1{\sqrt{ax+b}}dx=\frac2a\sqrt{ax+b}+C\;,\;\;a,b,C\;\text{constants}$$