Question

$\int_4^{18} \frac{1}{(x+2)\sqrt{x-3}}dx = $

• A. $\frac{\pi}{6\sqrt{5}}$
• B. $\frac{\pi}{6}$
• C. $\frac{\pi}{3}$
• D. $\frac{\pi}{3\sqrt{5}}$
• E. None of these

Hint:

For integrals involving $\sqrt{ax+b}$, the substitution $u^2=ax+b$ is almost always the best first step. It transforms the integrand from an irrational function to a rational function, which can then be handled with standard techniques like partial fractions or inverse trigonometric forms.

Answer 1

Answer: (E)

Step 1: Substitution Let \( u = \sqrt{x-3} \implies u^2 = x-3 \implies x = u^2+3 \). \( dx = 2u du \). Limits: \( x=8 \implies u=\sqrt{5} \). \( x=18 \implies u=\sqrt{15} \).
Step 2: Transform the Integral Denominator term \( x+2 = (u^2+3) + 2 = u^2+5 \). \[ I = \int_{\sqrt{5}}^{\sqrt{15}} \frac{2u du}{(u^2+5) \cdot u} = 2 \int_{\sqrt{5}}^{\sqrt{15}} \frac{du}{u^2+5} \] \[ I = 2 \int_{\sqrt{5}}^{\sqrt{15}} \frac{du}{u^2+(\sqrt{5})^2} \]
Step 3: Integrate and Evaluate \( I = 2 \cdot \frac{1}{\sqrt{5}} [ \tan^{-1}(u/\sqrt{5}) ]_{\sqrt{5}}^{\sqrt{15}} \). Upper limit: \( \tan^{-1}(\sqrt{15}/\sqrt{5}) = \tan^{-1}(\sqrt{3}) = \pi/3 \). Lower limit: \( \tan^{-1}(\sqrt{5}/\sqrt{5}) = \tan^{-1}(1) = \pi/4 \). \[ I = \frac{2}{\sqrt{5}} \left( \frac{\pi}{3} - \frac{\pi}{4} \right) = \frac{2}{\sqrt{5}} \left( \frac{\pi}{12} \right) = \frac{\pi}{6\sqrt{5}} \]