Key Concepts and Formulas

• Indefinite Integration: Finding a function whose derivative is the given integrand.
• Substitution Method: $\int f(g(x))g'(x) dx = \int f(u) du$, where $u = g(x)$.
• Algebraic manipulation and simplification of expressions.

Step-by-Step Solution

Step 1: Rewrite the integral

We begin by rewriting the given integral to make it easier to work with. $I = \int \frac{dx}{(1 + \sqrt{x})\sqrt{x - x^2}} = \int \frac{dx}{(1 + \sqrt{x})\sqrt{x}\sqrt{1 - x}}$

Step 2: First substitution: $1 + \sqrt{x} = t$

Let's substitute $1 + \sqrt{x} = t$. Then, differentiating both sides gives: $\frac{1}{2\sqrt{x}} dx = dt$ $dx = 2\sqrt{x} dt = 2(t-1) dt$ Substituting these into the integral, we get: $I = \int \frac{2(t-1) dt}{t\sqrt{(t-1)(1 - (t-1)^2)}} = \int \frac{2(t-1) dt}{t\sqrt{(t-1)(1 - (t^2 - 2t + 1))}} = \int \frac{2(t-1) dt}{t\sqrt{(t-1)(2t - t^2)}} = \int \frac{2(t-1) dt}{t\sqrt{t(t-1)(2 - t)}}$ $I = \int \frac{2(t-1) dt}{t \sqrt{t}\sqrt{(t-1)(2-t)}} = \int \frac{2(t-1) dt}{t \sqrt{2t - t^2}}$ $I = \int \frac{2(t-1) dt}{t \sqrt{2t - t^2}} = 2\int \frac{(t-1)}{t\sqrt{2t - t^2}} dt$ This doesn't seem to simplify the integral, so let's go back to the original substitution and rewrite the integral as: $I = \int \frac{dx}{(1 + \sqrt{x})\sqrt{x}\sqrt{1 - x}}$ Using the substitution $\frac{1}{2\sqrt{x}} dx = dt$, we have $dx = 2\sqrt{x} dt = 2(t-1)dt$. Substituting this into the original integral gives $I = \int \frac{2(t-1)dt}{t\sqrt{(t-1)^2(1 - (t-1)^2)}} = \int \frac{2(t-1)dt}{t(t-1)\sqrt{1 - (t^2 - 2t + 1)}} = \int \frac{2dt}{t\sqrt{2t - t^2}}$

Step 3: Second substitution: $t = \frac{1}{z}$

Let's substitute $t = \frac{1}{z}$. Then, $dt = -\frac{1}{z^2} dz$. Substituting these into the integral, we get: $I = 2\int \frac{-\frac{1}{z^2} dz}{\frac{1}{z}\sqrt{\frac{2}{z} - \frac{1}{z^2}}} = 2\int \frac{-\frac{1}{z^2} dz}{\frac{1}{z}\sqrt{\frac{2z - 1}{z^2}}} = 2\int \frac{-\frac{1}{z^2} dz}{\frac{1}{z} \cdot \frac{\sqrt{2z - 1}}{z}} = 2\int \frac{-dz}{\sqrt{2z - 1}}$

Step 4: Evaluate the simplified integral

$I = -2 \int \frac{dz}{\sqrt{2z - 1}}$ Let $u = 2z - 1$. Then $du = 2dz$, so $dz = \frac{1}{2}du$. $I = -2 \int \frac{\frac{1}{2}du}{\sqrt{u}} = -\int u^{-1/2} du = -\frac{u^{1/2}}{1/2} + C = -2\sqrt{u} + C = -2\sqrt{2z - 1} + C$

Step 5: Back-substitute to find the result in terms of $x$

Now, we substitute back to express the result in terms of $x$. First, we have $z = \frac{1}{t}$, so $I = -2\sqrt{\frac{2}{t} - 1} + C = -2\sqrt{\frac{2 - t}{t}} + C$ Since $t = 1 + \sqrt{x}$, we have $I = -2\sqrt{\frac{2 - (1 + \sqrt{x})}{1 + \sqrt{x}}} + C = -2\sqrt{\frac{1 - \sqrt{x}}{1 + \sqrt{x}}} + C$

Common Mistakes & Tips

• Be careful with algebraic manipulations, especially when dealing with square roots.
• Choosing the right substitution is crucial. If the initial substitution doesn't simplify the integral, try a different one.
• Remember to back-substitute to express the final answer in terms of the original variable.

Summary

We started by rewriting the integral. Then, we used the substitution $1 + \sqrt{x} = t$, followed by $t = \frac{1}{z}$. After these substitutions, we were able to evaluate the simplified integral and back-substitute to obtain the final result in terms of $x$. The final integral is $-2\sqrt{\frac{1 - \sqrt{x}}{1 + \sqrt{x}}} + C$.

Final Answer

The final answer is \boxed{- 2\sqrt {{{1 - \sqrt x } \over {1 + \sqrt x }}} + C}, which corresponds to option (B).