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:
This integral can be solved using the substitution method, as the derivative of the inner function is present in the integrand.
Key Formula or Approach:
Let \( u = \sqrt{x} + 1 \). Then differentiate to find $dx$ in terms of $du$.

Step 2: Detailed Explanation:
Let \( u = \sqrt{x} + 1 \).
Differentiating with respect to $x$:
\[ \frac{du}{dx} = \frac{1}{2\sqrt{x}} \implies \frac{1}{\sqrt{x}} dx = 2 du \]
Substitute these into the integral:
\[ I = \int \sqrt{u} \cdot 2 du = 2 \int u^{1/2} du \]
Integrate using the power rule:
\[ I = 2 \left( \frac{u^{3/2}}{3/2} \right) + C = 2 \cdot \frac{2}{3} u^{3/2} + C = \frac{4}{3} u^{3/2} + C \]
Substitute back \( u = \sqrt{x} + 1 \):
\[ I = \frac{4}{3} (\sqrt{x} + 1)^{3/2} + C \]

Step 3: Final Answer:
The integral is \( \frac{4}{3} (\sqrt{x} + 1)^{3/2} + C \).