If int frac{x^3 , dx}{sqrt{1+x^2}} = a(1+x^2) | vedclass

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(B) Let $I = \int \frac{x^3 \, dx}{\sqrt{1+x^2}}$.Put $1+x^2 = t^2$,then $2x \, dx = 2t \, dt$,which implies $x \, dx = t \, dt$.Also,$x^2 = t^2 - 1$.

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$-1$

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(B) Let $I = \int \frac{x^3 \, dx}{\sqrt{1+x^2}}$.Put $1+x^2 = t^2$,then $2x \, dx = 2t \, dt$,which implies $x \, dx = t \, dt$.Also,$x^2 = t^2 - 1$.Substituting these into the integral:$I = \int \frac{(t^2-1) \cdot t \, dt}{t} = \int (t^2 - 1) \, dt$.Integrating with respect to $t$:$I = \frac{t^3}{3} - t + c$.Substituting $t = \sqrt{1+x^2}$ back:$I = \frac{(1+x^2)^{3/2}}{3} - \sqrt{1+x^2} + c = \frac{1}{3}(1+x^2)\sqrt{1+x^2} - 1\sqrt{1+x^2} + c$.Comparing this with the given form $a(1+x^2)\sqrt{1+x^2} + b\sqrt{1+x^2} + c$,we get $a = \frac{1}{3}$ and $b = -1$.Therefore,$3ab = 3 	imes \frac{1}{3} 	imes (-1) = -1$.

If $\int \frac{x^3 \, dx}{\sqrt{1+x^2}} = a(1+x^2) \sqrt{1+x^2} + b \sqrt{1+x^2} + c$ (where $c$ is a constant of integration),then the value of $3ab$ is

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