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

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Question

(A) Let $t = \log _e(x+\sqrt{1+x^2})$.Then, $dt = \frac{1}{x+\sqrt{1+x^2}} \cdot (1 + \frac{2x}{2\sqrt{1+x^2}}) dx = \frac{1}{x+\sqrt{1+x^2}} \cdot \f

Vedclass · Accepted Answer

$f(x) = \frac{x^2}{2}, g(x) = \log _e(x+\sqrt{1+x^2})$

Vedclass · Answer

(A) Let $t = \log _e(x+\sqrt{1+x^2})$.Then, $dt = \frac{1}{x+\sqrt{1+x^2}} \cdot (1 + \frac{2x}{2\sqrt{1+x^2}}) dx = \frac{1}{x+\sqrt{1+x^2}} \cdot \frac{\sqrt{1+x^2}+x}{\sqrt{1+x^2}} dx = \frac{dx}{\sqrt{1+x^2}}$.Substituting this into the integral, we get:$I = \int t \, dt = \frac{t^2}{2} + c$.Substituting $t$ back, we get $I = \frac{(\log _e(x+\sqrt{1+x^2}))^2}{2} + c$.Comparing this with $f(g(x)) + c$, we have $f(x) = \frac{x^2}{2}$ and $g(x) = \log _e(x+\sqrt{1+x^2})$.

If $\int \frac{\log _e(x+\sqrt{1+x^2})}{\sqrt{1+x^2}} dx = f(g(x)) + c$, then:

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