Question

$\int \frac{dx}{(x-1)\sqrt{x+2}} =$

• A. $\frac{2}{\sqrt{3}} \log \left| \frac{\sqrt{x+2} + \sqrt{3}}{\sqrt{x+2} - \sqrt{3}} \right| + c$
• B. $-\frac{1}{\sqrt{3}} \log \left| \frac{\sqrt{x+2} - \sqrt{3}}{\sqrt{x+2} + \sqrt{3}} \right| + c$
• C. $\frac{1}{\sqrt{3}} \log \left| \frac{\sqrt{x+2} + \sqrt{3}}{\sqrt{x+2} - \sqrt{3}} \right| + c$
• D. $\frac{1}{\sqrt{3}} \log \left| \frac{\sqrt{x+2} - \sqrt{3}}{\sqrt{x+2} + \sqrt{3}} \right| + c$

Hint:

When the integrand involves $\sqrt{ax+b}$, a useful substitution is $u = \sqrt{ax+b}$.

Answer 1

The Correct Option is D

Step 1: Use the substitution $u = \sqrt{x+2$.}
Let $u = \sqrt{x+2}$, which implies $u^2 = x+2$, so $x = u^2 - 2$. Then, $dx = 2u \, du$.
Step 2: Substitute into the integral.
$$ \int \frac{dx}{(x-1)\sqrt{x+2}} = \int \frac{2u \, du}{(u^2 - 2 - 1)u} = \int \frac{2 \, du}{u^2 - 3} $$
Step 3: Evaluate the integral.
$$ \int \frac{2 \, du}{u^2 - (\sqrt{3})^2} = 2 \cdot \frac{1}{2\sqrt{3}} \log \left| \frac{u - \sqrt{3}}{u + \sqrt{3}} \right| + C = \frac{1}{\sqrt{3}} \log \left| \frac{u - \sqrt{3}}{u + \sqrt{3}} \right| + C $$
Step 4: Substitute back $u = \sqrt{x+2$.}
$$ \frac{1}{\sqrt{3}} \log \left| \frac{\sqrt{x+2} - \sqrt{3}}{\sqrt{x+2} + \sqrt{3}} \right| + C $$
Thus, the integral is $ \boxed{\frac{1}{\sqrt{3}} \log \left| \frac{\sqrt{x+2} - \sqrt{3}}{\sqrt{x+2} + \sqrt{3}} \right| + c} $.