Q.Evaluate the definite integral $\int_{0}^{1}\frac{dx}{\sqrt{1+x}-\sqrt{x}}$

The denominator $\sqrt{1+x}-\sqrt{x}$ never vanishes on $[0,1]$, so this is an ordinary definite integral. Rationalize it.

Multiply top and bottom by the conjugate $\sqrt{1+x}+\sqrt{x}$:

So …

First, is it improper?

At $x=0$ the denominator is $\sqrt{1}-\sqrt{0}=1$, and it stays positive across $[0,1]$, so there is no blow-up — this is a perfectly ordinary integral. The only difficulty is cosmetic: a difference of square roots on the bottom.

Rationalize the denominator

Whenever $\sqrt a-\sqrt b$ sits underneath, multiply top and bottom by the conjugate $\sqrt a+\sqrt b$, because $(\sqrt a-\sqrt b)(\sqrt a+\sqrt b)=a-b$:

The denominator becomes $1$, so the integral is now a sum of two easy power integrals.

Integrate each piece

Using $\int u^{1/2}\,du=\tfrac23u^{3/2}$: …

Method: Rationalise the difference of surds in the denominator

Use this for $\dfrac{1}{\sqrt{P}-\sqrt{Q}}$: multiply by the conjugate so the denominator becomes $P-Q$, leaving elementary power integrals.

Steps

Step 1: Multiply by the conjugate.

Step 2: Confirm the integral is proper. …

Common Mistakes

Mistake 1: Thinking the integral is improper.

Why it's wrong: at $x=0$ the denominator is $\sqrt1-\sqrt0=1\neq0$, and it stays positive on $[0,1]$; there is no singularity. Correct approach: treat it as an ordinary integral.

Mistake 2: Not rationalising.

Why it's wrong: leaving $\dfrac{1}{\sqrt{1+x}-\sqrt x}$ is hard to integrate; the conjugate turns it into $\sqrt{1+x}+\sqrt x$. Correct approach: multiply by the conjugate. …