Question:
Evaluate: $\int \frac{1}{\sqrt{x+3}-\sqrt{x+2}} d x$
Solution:
Let $I=\int \frac{1}{\sqrt{x+3}-\sqrt{x+2}} \mathrm{dx}$
$=\int \frac{1}{\sqrt{x+3}-\sqrt{x+2}} d x$
Now, Multiply with the conjugate
$=\int \frac{1}{\sqrt{x+3}-\sqrt{x+2}} \times \frac{\sqrt{x+3}+\sqrt{x+2}}{\sqrt{x+3}+\sqrt{x+2}} d x$
$=\int \frac{\sqrt{x+3}+\sqrt{x+2}}{(\sqrt{x+3})^{2}-(\sqrt{x+2})^{2}} d x$
$=\int \frac{\sqrt{x+3}+\sqrt{x+2}}{x+3-x-2} d x$
$=\int(\mathrm{x}+3)^{\frac{1}{2}}+(\mathrm{x}+2)^{\frac{1}{2}} \mathrm{dx}$
$=\frac{(x+3)^{\frac{3}{2}}}{\frac{3}{2}}+\frac{(x+2)^{\frac{3}{2}}}{\frac{3}{2}}$
Hence, $I=\frac{2}{3}(x+3)^{\frac{3}{2}}+\frac{2}{3}(x+2)^{\frac{3}{2}}+C$