Question

\[ \int\limits_0^1 \frac{1}{2+\sqrt{x}} d x= \]

• A. \(2\mathrm{log}(\frac{2\mathrm{e}}{3})\)
• B. \(2\mathrm{log}(\frac{4\mathrm{e}}{9})\)
• C. \(\mathrm{log}(\frac{2\mathrm{e}}{3})\)
• D. \(\mathrm{log}(\frac{4e}{9})\)

Hint:

Substitute t = sqrt x to turn it into a rational integral.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept
Put \(t=\sqrt x\), so \(x=t^2\) and \(dx=2t\,dt\). The limits stay \(0\) to \(1\).

Step 2: Rewrite
\(I=\int_0^1\frac{2t}{2+t}dt=2\int_0^1\left(1-\frac2{2+t}\right)dt\).

Step 3: Integrate
\(I=2\left[t-2\ln(2+t)\right]_0^1=2\left[1-2\ln3+2\ln2\right]\).

Step 4: Simplify
\(I=2\left[\ln e+\ln\frac49\right]=2\ln\left(\frac{4e}9\right)\), option (B).

Final Answer:
The integral equals \(2\log\left(\frac{4e}{9}\right)\). \[ \boxed{\text{(B) }2\log\left(\dfrac{4\mathrm{e}}{9}\right)} \]