Evaluate:

Question:

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

Solution:

Let $I=\int \frac{1}{\sqrt{x+1}+\sqrt{x}} d x$

$=\int \frac{1}{\sqrt{x+1}+\sqrt{x}} d x$

Now Multiply with the conjugate, we get

$=\int \frac{1}{\sqrt{x+1}+\sqrt{x}} \cdot \frac{\sqrt{x+1}-\sqrt{x}}{\sqrt{x+1}-\sqrt{x}} d x$

$=\int \frac{\sqrt{x+1}-\sqrt{x}}{x+1-x} d x$

$=\int \sqrt{x+1}-\sqrt{x} d x$

$=\int(x+1)^{\frac{1}{2}}-x^{\frac{1}{2}}$

$=\frac{(x+1)^{\frac{3}{2}}}{\frac{3}{2}}-\frac{x^{\frac{3}{2}}}{\frac{3}{2}}$

Hence $I=\frac{2}{3}(x+1)^{\frac{3}{2}}-\frac{2}{3}(x)^{\frac{3}{2}}+C$

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