int frac{log (x + sqrt {1 + x^2})}{sqrt {1 + | vedclass

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Question

(A) Let $I = \int \frac{\log (x + \sqrt {1 + x^2})}{\sqrt {1 + x^2}} \, dx$. Substitute $t = \log (x + \sqrt {1 + x^2})$. Differentiating both sides w

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$\frac{1}{2}[\log (x + \sqrt {1 + x^2})]^2 + c$

Vedclass · Answer

(A) Let $I = \int \frac{\log (x + \sqrt {1 + x^2})}{\sqrt {1 + x^2}} \, dx$. Substitute $t = \log (x + \sqrt {1 + x^2})$. Differentiating both sides with respect to $x$,we get: $\frac{dt}{dx} = \frac{1}{x + \sqrt {1 + x^2}} \cdot (1 + \frac{1}{2\sqrt {1 + x^2}} \cdot 2x) = \frac{1}{x + \sqrt {1 + x^2}} \cdot (1 + \frac{x}{\sqrt {1 + x^2}}) = \frac{1}{x + \sqrt {1 + x^2}} \cdot \frac{\sqrt {1 + x^2} + x}{\sqrt {1 + x^2}} = \frac{1}{\sqrt {1 + x^2}}$. Thus,$dt = \frac{dx}{\sqrt {1 + x^2}}$. Substituting these into the integral: $I = \int t \, dt = \frac{t^2}{2} + c$. Replacing $t$ with its original value: $I = \frac{1}{2}[\log (x + \sqrt {1 + x^2})]^2 + c$.

$\int \frac{\log (x + \sqrt {1 + x^2})}{\sqrt {1 + x^2}} \, dx = $

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