int sqrt{x+sqrt{x^2+2}} , dx = | vedclass

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(D) Let $I = \int \sqrt{x+\sqrt{x^2+2}} \, dx$. We know that $(\sqrt{x^2+2}+x)(\sqrt{x^2+2}-x) = (x^2+2) - x^2 = 2$. Let $t = x+\sqrt{x^2+2}$. Then $\

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$\frac{(x+\sqrt{x^2+2})^2-6}{3\sqrt{x+\sqrt{x^2+2}}}+C$

Vedclass · Answer

(D) Let $I = \int \sqrt{x+\sqrt{x^2+2}} \, dx$. We know that $(\sqrt{x^2+2}+x)(\sqrt{x^2+2}-x) = (x^2+2) - x^2 = 2$. Let $t = x+\sqrt{x^2+2}$. Then $\sqrt{x^2+2}-x = \frac{2}{t}$. Adding these two equations,we get $2\sqrt{x^2+2} = t + \frac{2}{t}$,so $\sqrt{x^2+2} = \frac{t^2+2}{2t}$. Subtracting the equations,we get $2x = t - \frac{2}{t}$,so $x = \frac{t^2-2}{2t}$. Differentiating $x$ with respect to $t$,we get $dx = \frac{d}{dt}(\frac{t}{2} - \frac{1}{t}) \, dt = (\frac{1}{2} + \frac{1}{t^2}) \, dt = \frac{t^2+2}{2t^2} \, dt$. Substituting these into the integral: $I = \int \sqrt{t} \cdot \frac{t^2+2}{2t^2} \, dt = \frac{1}{2} \int (t^{1/2} + 2t^{-3/2}) \, dt$. $I = \frac{1}{2} [\frac{t^{3/2}}{3/2} + 2 \cdot \frac{t^{-1/2}}{-1/2}] + C$. $I = \frac{1}{2} [\frac{2}{3}t^{3/2} - 4t^{-1/2}] + C = \frac{1}{3}t^{3/2} - 2t^{-1/2} + C$. $I = \frac{t^2-6}{3\sqrt{t}} + C$. Substituting $t = x+\sqrt{x^2+2}$ back,we get $I = \frac{(x+\sqrt{x^2+2})^2-6}{3\sqrt{x+\sqrt{x^2+2}}} + C$.

$\int \sqrt{x+\sqrt{x^2+2}} \, dx =$

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