int_0^1 frac{dx}{(3x+2)+sqrt{3x+2}} = . . . | vedclass

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(C) Let $I = \int_0^1 \frac{dx}{(3x+2)+\sqrt{3x+2}}$.Substitute $u = \sqrt{3x+2}$. Then $u^2 = 3x+2$,so $2u du = 3 dx$,which means $dx = \frac{2}{3} u

Vedclass · Accepted Answer

$\frac{2}{3} \log \left|\frac{\sqrt{5}+1}{\sqrt{2}+1}\right|$

Vedclass · Answer

(C) Let $I = \int_0^1 \frac{dx}{(3x+2)+\sqrt{3x+2}}$.Substitute $u = \sqrt{3x+2}$. Then $u^2 = 3x+2$,so $2u du = 3 dx$,which means $dx = \frac{2}{3} u du$.When $x=0$,$u = \sqrt{2}$. When $x=1$,$u = \sqrt{5}$.Substituting these into the integral:$I = \int_{\sqrt{2}}^{\sqrt{5}} \frac{\frac{2}{3} u du}{u^2+u} = \frac{2}{3} \int_{\sqrt{2}}^{\sqrt{5}} \frac{u du}{u(u+1)} = \frac{2}{3} \int_{\sqrt{2}}^{\sqrt{5}} \frac{du}{u+1}$.Integrating,we get $I = \frac{2}{3} [\log |u+1|]_{\sqrt{2}}^{\sqrt{5}}$.$I = \frac{2}{3} (\log |\sqrt{5}+1| - \log |\sqrt{2}+1|) = \frac{2}{3} \log \left|\frac{\sqrt{5}+1}{\sqrt{2}+1}\right|$.Thus,the correct option is $C$.

$\int_0^1 \frac{dx}{(3x+2)+\sqrt{3x+2}} = $ . . . . . . .

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