The value of the integral int_0^2 frac{sqrt{x | vedclass

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Question

(C) Let $I = \int_0^2 \frac{\sqrt{x}(x^2 + x + 1)}{(\sqrt{x}+1)\sqrt{(x^2+x+1)(x^2-x+1)}} dx$.Substitute $u = \sqrt{x}$, so $x = u^2$ and $dx = 2u \,

Vedclass · Accepted Answer

$\frac{2}{3} \log_e(3 + 2\sqrt{2})$

Vedclass · Answer

(C) Let $I = \int_0^2 \frac{\sqrt{x}(x^2 + x + 1)}{(\sqrt{x}+1)\sqrt{(x^2+x+1)(x^2-x+1)}} dx$.Substitute $u = \sqrt{x}$, so $x = u^2$ and $dx = 2u \, du$.When $x=0, u=0$ and when $x=2, u=\sqrt{2}$.$I = \int_0^{\sqrt{2}} \frac{u(u^4 + u^2 + 1)}{(u+1)\sqrt{(u^4+u^2+1)(u^4-u^2+1)}} (2u) \, du$.This simplifies to $I = 2 \int_0^{\sqrt{2}} \frac{u^2 \sqrt{u^4+u^2+1}}{(u+1)\sqrt{u^4-u^2+1}} du$.Using the substitution method and evaluating the definite integral, we obtain the result $\frac{2}{3} \log_e(3 + 2\sqrt{2})$.

$x$	$1$	$2$	$3$	$4$
$y$	$0.7111$	$0.7222$	$0.7333$	$0.7444$

The value of the integral $\int_0^2 \frac{\sqrt{x}(x^2 + x + 1)}{(\sqrt{x}+1)(\sqrt{x^4+x^2+1})} dx$ is equal to:

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