int {frac{{xdx}}{{sqrt {1 + {x^2} + sqrt {{{l | vedclass

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(D) Let $I = \int \frac{xdx}{\sqrt{1 + x^2 + (1 + x^2)^{3/2}}}$.Substitute $1 + x^2 = t^2$,then $2xdx = 2tdt$,which implies $xdx = tdt$.Substituting t

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

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(D) Let $I = \int \frac{xdx}{\sqrt{1 + x^2 + (1 + x^2)^{3/2}}}$.Substitute $1 + x^2 = t^2$,then $2xdx = 2tdt$,which implies $xdx = tdt$.Substituting these into the integral:$I = \int \frac{tdt}{\sqrt{t^2 + (t^2)^{3/2}}} = \int \frac{tdt}{\sqrt{t^2 + t^3}}$.Since $t > 0$,$\sqrt{t^2 + t^3} = t\sqrt{1 + t}$.$I = \int \frac{tdt}{t\sqrt{1 + t}} = \int \frac{dt}{\sqrt{1 + t}}$.Integrating with respect to $t$:$I = 2\sqrt{1 + t} + C$.Substituting back $t = \sqrt{1 + x^2}$:$I = 2\sqrt{1 + \sqrt{1 + x^2}} + C$.

$\int {\frac{{xdx}}{{\sqrt {1 + {x^2} + \sqrt {{{\left( {1 + {x^2}} \right)}^3}} } }}} $ is equal to (where $C$ denotes constant of integration)

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