int_{8}^{18} frac{1}{(x+2) sqrt{x-3}} , dx = | vedclass

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Question

(A) Let $I = \int_{8}^{18} \frac{1}{(x+2) \sqrt{x-3}} \, dx$. Substitute $t = \sqrt{x-3}$,so $t^2 = x-3$,which implies $x = t^2+3$ and $dx = 2t \, dt$

Vedclass · Accepted Answer

$\frac{\pi}{6 \sqrt{5}}$

Vedclass · Answer

(A) Let $I = \int_{8}^{18} \frac{1}{(x+2) \sqrt{x-3}} \, dx$. Substitute $t = \sqrt{x-3}$,so $t^2 = x-3$,which implies $x = t^2+3$ and $dx = 2t \, dt$. When $x = 8$,$t = \sqrt{8-3} = \sqrt{5}$. When $x = 18$,$t = \sqrt{18-3} = \sqrt{15}$. Substituting these into the integral: $I = \int_{\sqrt{5}}^{\sqrt{15}} \frac{2t \, dt}{(t^2+3+2)t} = \int_{\sqrt{5}}^{\sqrt{15}} \frac{2 \, dt}{t^2+5}$. Using the formula $\int \frac{dx}{x^2+a^2} = \frac{1}{a} 	an^{-1}(\frac{x}{a}) + C$: $I = 2 \left[ \frac{1}{\sqrt{5}} 	an^{-1}(\frac{t}{\sqrt{5}}) ight]_{\sqrt{5}}^{\sqrt{15}}$. $I = \frac{2}{\sqrt{5}} \left[ 	an^{-1}(\frac{\sqrt{15}}{\sqrt{5}}) - 	an^{-1}(\frac{\sqrt{5}}{\sqrt{5}}) ight]$. $I = \frac{2}{\sqrt{5}} \left[ 	an^{-1}(\sqrt{3}) - 	an^{-1}(1) ight]$. $I = \frac{2}{\sqrt{5}} \left[ \frac{\pi}{3} - \frac{\pi}{4} ight] = \frac{2}{\sqrt{5}} \left[ \frac{\pi}{12} ight] = \frac{\pi}{6 \sqrt{5}}$.

$\int_{8}^{18} \frac{1}{(x+2) \sqrt{x-3}} \, dx = $

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