Domain of definition of the function f(x) = i | vedclass

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(D) The function is defined as $f(x) = \int\limits_0^x \frac{dt}{\sqrt{x^2 + t^2}}$.Let $t = xu$,then $dt = x du$.When $t = 0$,$u = 0$. When $t = x$,$

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$R - \{0\}$

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(D) The function is defined as $f(x) = \int\limits_0^x \frac{dt}{\sqrt{x^2 + t^2}}$.Let $t = xu$,then $dt = x du$.When $t = 0$,$u = 0$. When $t = x$,$u = 1$.Substituting these into the integral,we get $f(x) = \int\limits_0^1 \frac{x du}{\sqrt{x^2 + x^2 u^2}} = \int\limits_0^1 \frac{x du}{|x| \sqrt{1 + u^2}}$.If $x > 0$,then $f(x) = \int\limits_0^1 \frac{du}{\sqrt{1 + u^2}} = [\ln(u + \sqrt{1 + u^2})]_0^1 = \ln(1 + \sqrt{2})$.If $x If $x = 0$,the integral becomes $\int\limits_0^0 \frac{dt}{t}$,which is undefined.Thus,the function is defined for all $x 
eq 0$,which is $R - \{0\}$.

Domain of definition of the function $f(x) = \int\limits_0^x \frac{dt}{\sqrt{x^2 + t^2}}$ is

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