Question

Evaluate $\displaystyle \int \frac{dx}{1+\sqrt{x}}$

• A. $2\sqrt{x} - 2\log|1+\sqrt{x}| + c$
• B. $\sqrt{x} + \log|1+\sqrt{x}| + c$
• C. $2\sqrt{x} + \log|1+\sqrt{x}| + c$
• D. $\sqrt{x} - \log|1+\sqrt{x}| + c$

Hint:

When radicals appear in the denominator, substitution often simplifies the integral.

Answer 1

The Correct Option is A

Step 1: Using substitution.
Let $\sqrt{x} = t$, so that $x = t^2$ and $dx = 2t\,dt$.
Step 2: Transforming the integral.
\[ \int \frac{dx}{1+\sqrt{x}} = \int \frac{2t}{1+t}\,dt \]
Step 3: Simplifying the integrand.
\[ \frac{2t}{1+t} = 2\left(1 - \frac{1}{1+t}\right) \]
Step 4: Integrating.
\[ \int 2\,dt - \int \frac{2}{1+t}\,dt = 2t - 2\log|1+t| + c \]
Step 5: Substituting back.
\[ = 2\sqrt{x} - 2\log|1+\sqrt{x}| + c \]