Q.Integrate the function $\sqrt{\frac{1-\sqrt{x}}{1+\sqrt{x}}}$

Rationalise by multiplying inside the root by $\dfrac{1-\sqrt x}{1-\sqrt x}$ (for $0<x<1$):

First term: $\displaystyle\int\frac{dx}{\sqrt{1-x}}=-2\sqrt{1-x}.$ …

1. Kill the nested root

Multiply numerator and denominator inside the square root by $1-\sqrt x$ (valid for $0<x<1$, where $1-\sqrt x\ge0$):

Split it into two manageable pieces:

2. First integral

3. Second integral

Put $x=\sin^2\theta$, so $dx=2\sin\theta\cos\theta\,d\theta$, $\sqrt x=\sin\theta$, $\sqrt{1-x}=\cos\theta$:

$\int\frac{\sqrt x}{\sqrt{1-x}}\,dx=\int\frac{\sin\theta}{\cos\theta}\,2\sin\theta\cos\theta\,d\theta=2\int\sin^2\theta\,d\theta=\int(1-\cos2\theta)\,d\theta=\theta-\sin\theta\cos\theta.$ …

Method: Rationalise a nested surd, then split and use a trig substitution

Use this for $\sqrt{\dfrac{1-\sqrt x}{1+\sqrt x}}$-type integrands: multiply inside the root by the conjugate to remove the nesting, then handle each resulting piece separately.

Steps

Step 1: Rationalise inside the square root.

Multiply numerator and denominator inside the root by $1-\sqrt x$ (valid where $1-\sqrt x\ge0$), giving $\dfrac{1-\sqrt x}{\sqrt{1-x}}$.

Step 2: Split into manageable pieces.

$\frac{1-\sqrt x}{\sqrt{1-x}}=\frac{1}{\sqrt{1-x}}-\frac{\sqrt x}{\sqrt{1-x}}.$ …

Common Mistakes

Mistake 1: Ignoring the domain when rationalising.

Why it's wrong: multiplying inside the root by $1-\sqrt x$ needs $1-\sqrt x\ge0$, i.e. $0<x<1$. Correct approach: state the domain so the surd manipulation is valid.

Mistake 2: Mishandling the split $\dfrac{1-\sqrt x}{\sqrt{1-x}}$. …