The value of \( \int\displaystyle\frac{dx}{1+\sqrt{x}} \) is
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → \( 2\sqrt{x}-2\log_e\left|1+\sqrt{x}\right|+C \) : where \( C \) is an arbitrary constant
Let \( \sqrt{x}=t \). Then \( x=t^2 \) and \( dx=2t\,dt \).
\( \displaystyle \int\frac{dx}{1+\sqrt{x}}=\int\frac{2t}{1+t}\,dt \)
\( \displaystyle =2\int\left(1-\frac{1}{1+t}\right)dt \)
\( \displaystyle =2t-2\log_e|1+t|+C \)
\( \displaystyle =2\sqrt{x}-2\log_e\left|1+\sqrt{x}\right|+C \)
Why the other options are incorrect
Option (1) \( \sqrt{x}\left(1+\sqrt{x}\right)+C \) : where \( C \) is an arbitrary constant — Incorrect. This equals \( \sqrt{x}+x+C \), whose derivative is \( \frac{1}{2\sqrt{x}}+1 \), not \( \frac{1}{1+\sqrt{x}} \).
Option (2) \( \sqrt{x}+\log_e\left|1+\sqrt{x}\right|+C \) : where \( C \) is an arbitrary constant — Incorrect. The factor 2 is missing and the log term has the wrong sign.
Option (3) \( \sqrt{x}-\log_e\left|1+\sqrt{x}\right|+C \) : where \( C \) is an arbitrary constant — Incorrect. This drops the factor 2 that comes from \( dx=2t\,dt \).