Question

\[ \int \frac{dx}{\sqrt{x+1}+\sqrt{x}}= \]

• A. \(\frac{2}{3}\left[(x+1)^{\frac{3}{2}}-x^{\frac{3}{2}}\right]+c\)
• B. \(\frac{2}{3}\left[(x+1)^{\frac{3}{2}}+x^{\frac{3}{2}}\right]+c\)
• C. \(\frac{3}{2}\left[(x+1)^{\frac{3}{2}}-x^{\frac{3}{2}}\right]+c\)
• D. \(\frac{3}{2}\left[(x+1)^{\frac{3}{2}}+x^{\frac{3}{2}}\right]+c\)

Hint:

For integrals with \(\sqrt{x+1}+\sqrt{x}\) in the denominator, rationalize using \(\sqrt{x+1}-\sqrt{x}\).

Answer 1

The Correct Option is A

Concept: When an expression has a denominator of the form: \[ \sqrt{x+1}+\sqrt{x} \] we rationalize it using the conjugate: \[ \sqrt{x+1}-\sqrt{x} \]

Step 1: Given integral: \[ \int \frac{dx}{\sqrt{x+1}+\sqrt{x}} \]

Step 2: Rationalize the denominator. \[ \frac{1}{\sqrt{x+1}+\sqrt{x}} \cdot \frac{\sqrt{x+1}-\sqrt{x}}{\sqrt{x+1}-\sqrt{x}} \] \[ = \frac{\sqrt{x+1}-\sqrt{x}}{(x+1)-x} \] \[ = \sqrt{x+1}-\sqrt{x} \]

Step 3: Therefore, the integral becomes: \[ \int \left(\sqrt{x+1}-\sqrt{x}\right)\,dx \] \[ = \int (x+1)^{\frac{1}{2}}\,dx-\int x^{\frac{1}{2}}\,dx \]

Step 4: Integrate each term. \[ \int (x+1)^{\frac{1}{2}}\,dx = \frac{(x+1)^{\frac{3}{2}}}{\frac{3}{2}} = \frac{2}{3}(x+1)^{\frac{3}{2}} \] \[ \int x^{\frac{1}{2}}\,dx = \frac{x^{\frac{3}{2}}}{\frac{3}{2}} = \frac{2}{3}x^{\frac{3}{2}} \]

Step 5: Combine the result. \[ \int \frac{dx}{\sqrt{x+1}+\sqrt{x}} = \frac{2}{3}(x+1)^{\frac{3}{2}}-\frac{2}{3}x^{\frac{3}{2}}+c \] \[ = \frac{2}{3}\left[(x+1)^{\frac{3}{2}}-x^{\frac{3}{2}}\right]+c \] Therefore, \[ \boxed{\frac{2}{3}\left[(x+1)^{\frac{3}{2}}-x^{\frac{3}{2}}\right]+c} \]