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

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(A) Let $t = \log \left(x+\sqrt{1+x^2}\right)$.Then,differentiating both sides with respect to $x$,we get:$\frac{dt}{dx} = \frac{1}{x+\sqrt{1+x^2}} \c

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$\log \left(x+\sqrt{1+x^2}\right)$

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(A) Let $t = \log \left(x+\sqrt{1+x^2}\right)$.Then,differentiating both sides with respect to $x$,we get:$\frac{dt}{dx} = \frac{1}{x+\sqrt{1+x^2}} \cdot \left(1 + \frac{1}{2\sqrt{1+x^2}} \cdot 2x\right) = \frac{1}{x+\sqrt{1+x^2}} \cdot \left(\frac{\sqrt{1+x^2}+x}{\sqrt{1+x^2}}\right) = \frac{1}{\sqrt{1+x^2}}$.Thus,$dt = \frac{dx}{\sqrt{1+x^2}}$.Substituting these into the integral:$\int \frac{\log \left(x+\sqrt{1+x^2}\right)}{\sqrt{1+x^2}} \,dx = \int t \,dt = \frac{t^2}{2} + C$.Comparing this with the given expression $\frac{1}{2}(g(x))^2 + C$,we have:$\frac{1}{2}(g(x))^2 = \frac{1}{2} \left(\log \left(x+\sqrt{1+x^2}\right)\right)^2$.Therefore,$g(x) = \log \left(x+\sqrt{1+x^2}\right)$.

$\int \frac{\log \left(x+\sqrt{1+x^2}\right)}{\sqrt{1+x^2}} \,dx = \frac{1}{2}(g(x))^2 + C$,(where $C$ is the constant of integration). Then $g(x) =$

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