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

Concept: Rationalizing Denominator — multiply numerator and denominator by the conjugate to simplify the integrand.

Step 1: Multiply numerator and denominator by $\sqrt{x+a} - \sqrt{x+b}$:

Step 2: Simplify the denominator:

So the integrand becomes

Step 3: Integrate term by term:

Using $\int (x+c)^{1/2}\,dx = \frac{2}{3}(x+c)^{3/2} + C$, we get

When you see a sum of square roots in the denominator, your first instinct should be to rationalize. The reason is simple: square roots are messy to integrate directly, but after rationalization, the denominator becomes a simple difference of the terms inside the roots — which is a constant. That turns a complicated-looking fraction into a clean difference of two power functions.

Let’s walk through it.

1. Rationalize the denominator. Multiply numerator and denominator by the conjugate $\sqrt{x+a} - \sqrt{x+b}$:

2. Simplify the denominator. The product $(\sqrt{x+a}+\sqrt{x+b})(\sqrt{x+a} - \sqrt{x+b})$ is of the form $(p+q)(p-q) = p^2 - q^2$. Here $p = \sqrt{x+a}$ and $q = \sqrt{x+b}$, so:

This is a constant — that’s the whole point. The integral becomes:

3. Integrate each square root. Each term is of the form $\sqrt{x+c} = (x+c)^{1/2}$. The power rule for integration gives:

So:

4. Combine the results. Putting it all together:

Method: Rationalising a sum of surds in the denominator

Use this whenever the denominator is $\sqrt{P}\pm\sqrt{Q}$: multiply by the conjugate so the denominator collapses to $P-Q$.

Steps

Step 1: Multiply numerator and denominator by the conjugate.

For $\dfrac{1}{\sqrt P+\sqrt Q}$ multiply by $\dfrac{\sqrt P-\sqrt Q}{\sqrt P-\sqrt Q}$. Using $(u+v)(u-v)=u^2-v^2$, the denominator becomes $P-Q$.

Step 2: Recognise the new denominator is a constant (or simpler).

When $P-Q$ is a constant (as with $(x+a)-(x+b)=a-b$), pull it straight out of the integral — it does not depend on $x$.

Step 3: Integrate the leftover power functions.

You are left with a difference of terms like $\sqrt{x+c}=(x+c)^{1/2}$; apply the power rule

Preserve the order of the two terms from the numerator so the final signs stay correct.

Common Mistakes

Mistake 1: Treating $a-b$ as a function of $x$.

Why it's wrong: after rationalising, the denominator is the constant $(x+a)-(x+b)=a-b$; trying to "integrate" it is meaningless. Correct approach: pull the constant $\tfrac{1}{a-b}$ outside the integral.

Mistake 2: Wrong power-rule antiderivative for $\sqrt{x+c}$.

Why it's wrong: $\int(x+c)^{1/2}\,dx=\tfrac{2}{3}(x+c)^{3/2}$, but students often write $(x+c)^{3/2}$ without the $\tfrac23$. Correct approach: divide by the new exponent $\tfrac32$, i.e. multiply by $\tfrac23$.

Mistake 3: Swapping the order of the two roots.

Why it's wrong: the numerator after rationalising is $\sqrt{x+a}-\sqrt{x+b}$, so the first term must be $(x+a)^{3/2}$; reversing them flips the sign. Correct approach: keep the conjugate's order.