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: Understanding the Concept
This is a definite integral involving an algebraic function with a square root. A standard technique for integrals of this form is to use a substitution that eliminates the square root.
Step 2: Key Formula or Approach
1. Use the substitution $u^2 = x-3$. This implies $x = u^2+3$ and $dx = 2u du$. 2. Change the limits of integration based on the substitution. 3. Rewrite the entire integral in terms of $u$. The integrand should simplify to a rational function. 4. Evaluate the resulting integral, which will likely be in the form of an inverse tangent.
Step 3: Detailed Explanation
Let $I = \int_4^{18} \frac{1}{(x+2)\sqrt{x-3}}dx$. 1. Substitution: Let $u^2 = x-3$. Then $\sqrt{x-3} = u$. Also, $x = u^2+3$, which means $dx = 2u du$. The term in the denominator is $x+2 = (u^2+3)+2 = u^2+5$. 2. Change the limits: Lower limit: When $x=4$, $u^2 = 4-3=1 \implies u=1$. Upper limit: When $x=18$, $u^2 = 18-3=15 \implies u=\sqrt{15}$. 3. Rewrite the integral: Substitute everything back into the integral: \[ I = \int_1^{\sqrt{15}} \frac{1}{(u^2+5) \cdot u} (2u du) \] The $u$ terms cancel out: \[ I = \int_1^{\sqrt{15}} \frac{2}{u^2+5} du = 2 \int_1^{\sqrt{15}} \frac{1}{u^2+(\sqrt{5})^2} du \] 4. Evaluate the integral: This is a standard inverse tangent integral of the form $\int \frac{1}{a^2+u^2}du = \frac{1}{a}\arctan\left(\frac{u}{a}\right)$. Here, $a=\sqrt{5}$. \[ I = 2 \left[ \frac{1}{\sqrt{5}} \arctan\left(\frac{u}{\sqrt{5}}\right) \right]_1^{\sqrt{15}} \] \[ I = \frac{2}{\sqrt{5}} \left[ \arctan\left(\frac{\sqrt{15}}{\sqrt{5}}\right) - \arctan\left(\frac{1}{\sqrt{5}}\right) \right] \] \[ I = \frac{2}{\sqrt{5}} \left[ \arctan(\sqrt{3}) - \arctan\left(\frac{1}{\sqrt{5}}\right) \right] \] We know that $\arctan(\sqrt{3}) = \frac{\pi}{3}$. \[ I = \frac{2}{\sqrt{5}} \left( \frac{\pi}{3} - \arctan\left(\frac{1}{\sqrt{5}}\right) \right) \] This is the exact value of the integral. However, it does not match any of the simple options given. The options are all constant multiples of $\pi$, suggesting that the arctangent terms should have evaluated to a simple fraction of $\pi$, which is not the case here. Step 4: Final Answer
The correct value of the integral is $\frac{2}{\sqrt{5}} \left( \frac{\pi}{3} - \arctan\left(\frac{1}{\sqrt{5}}\right) \right)$. Since this does not simplify to any of the given options, the question or options are flawed.