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

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(D) Let $I = \int \frac{\left(x+\sqrt{1+x^2}\right)^2}{\sqrt{1+x^2}} d x$. Substitute $t = x + \sqrt{1+x^2}$. Differentiating with respect to $x$, we

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

Vedclass · Answer

(D) Let $I = \int \frac{\left(x+\sqrt{1+x^2}\right)^2}{\sqrt{1+x^2}} d x$. Substitute $t = x + \sqrt{1+x^2}$. Differentiating with respect to $x$, we get $dt = \left(1 + \frac{2x}{2\sqrt{1+x^2}}\right) dx = \left(1 + \frac{x}{\sqrt{1+x^2}}\right) dx = \left(\frac{\sqrt{1+x^2} + x}{\sqrt{1+x^2}}\right) dx$. Thus, the integral becomes $I = \int t dt$. Integrating $t$ with respect to $t$, we get $I = \frac{t^2}{2} + C$. Substituting $t$ back, we get $I = \frac{\left(x+\sqrt{1+x^2}\right)^2}{2} + C$.

$\int \frac{\left(x+\sqrt{1+x^2}\right)^2}{\sqrt{1+x^2}} d x=$

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