int xsqrt{1 + x^2} , dx = | vedclass

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(D) To evaluate the integral $\int x\sqrt{1 + x^2} \, dx$,we use the method of substitution.Let $t = 1 + x^2$.Then,differentiating both sides with res

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$\frac{1}{3}(1 + x^2)^{3/2} + c$

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(D) To evaluate the integral $\int x\sqrt{1 + x^2} \, dx$,we use the method of substitution.Let $t = 1 + x^2$.Then,differentiating both sides with respect to $x$,we get $dt = 2x \, dx$,which implies $x \, dx = \frac{dt}{2}$.Substituting these into the integral,we get:$\int \sqrt{t} \cdot \frac{dt}{2} = \frac{1}{2} \int t^{1/2} \, dt$.Using the power rule for integration,$\int t^n \, dt = \frac{t^{n+1}}{n+1} + c$,we have:$\frac{1}{2} \cdot \frac{t^{3/2}}{3/2} + c = \frac{1}{2} \cdot \frac{2}{3} t^{3/2} + c = \frac{1}{3} t^{3/2} + c$.Finally,substituting $t = 1 + x^2$ back,we get $\frac{1}{3}(1 + x^2)^{3/2} + c$.

$\int x\sqrt{1 + x^2} \, dx = $

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