Q.Evaluate : $\int \dfrac{1}{\sqrt{x+2}-\sqrt{x+3}}\,dx$

Rationalise, then use $\int(x+a)^{1/2}dx=\tfrac23(x+a)^{3/2}$: answer $=-\tfrac23[(x+2)^{3/2}+(x+3)^{3/2}]+c$.

$I=\int\frac{1}{\sqrt{x+2}-\sqrt{x+3}}\,dx.$

Step 1 — rationalise by multiplying numerator and denominator by the conjugate $\sqrt{x+2}+\sqrt{x+3}$:

$\frac{1}{\sqrt{x+2}-\sqrt{x+3}}\cdot\frac{\sqrt{x+2}+\sqrt{x+3}}{\sqrt{x+2}+\sqrt{x+3}}=\frac{\sqrt{x+2}+\sqrt{x+3}}{(x+2)-(x+3)}=\frac{\sqrt{x+2}+\sqrt{x+3}}{-1}.$

So $I=-\displaystyle\int\left(\sqrt{x+2}+\sqrt{x+3}\right)dx.$

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