Question

$\int \frac{\sqrt{\sqrt{x} + 1}}{\sqrt{x}} dx = $}

• A. $\frac{4}{3} (\sqrt{x} + 1)^{3/2} + C$
• B. $\frac{2}{3} (\sqrt{x} + 1)^{3/2} + C$
• C. $\frac{4}{3} (\sqrt{x} + 1)^{3/4} + C$
• D. $\frac{1}{3} (\sqrt{x} + 1)^{3/2} + C$
• E. $\frac{3}{4} (\sqrt{x} + 1)^{3/2} + C$

Hint:

Whenever you see \( \sqrt{x} \) in a denominator and another \( \sqrt{x} \) inside another root, substitution is almost always the intended path. The \( 1/2 \) from the derivative often converts to a factor of 2 in the numerator.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
The integral involves composite root functions. A standard substitution method will simplify the integrand.
Step 2: Key Formula or Approach:
Let \(u\) equal the expression inside the outer square root.
\(u = \sqrt{x} + 1\).
Differentiate to find \(dx\) in terms of \(du\).
Step 3: Detailed Explanation:
Let \(u = \sqrt{x} + 1 = x^{1/2} + 1\).
Find \(du\):
\[ du = \frac{1}{2}x^{-1/2} dx = \frac{1}{2\sqrt{x}} dx \] Multiply by 2 to match the integrand's format:
\[ 2du = \frac{1}{\sqrt{x}} dx \] Substitute into the original integral:
\[ I = \int \sqrt{u} (2 du) = 2 \int u^{1/2} du \] Integrate using the power rule:
\[ I = 2 \left( \frac{u^{3/2}}{3/2} \right) + C = 2 \left( \frac{2}{3} \right) u^{3/2} + C \] \[ I = \frac{4}{3} u^{3/2} + C \] Substitute back \(u = \sqrt{x} + 1\):
\[ I = \frac{4}{3}(\sqrt{x} + 1)^{3/2} + C \] Step 4: Final Answer:
The result is \(\frac{4}{3}(\sqrt{x} + 1)^{\frac{3}{2}} + C\).