Q.Integrate the following function: $\frac{1}{\sqrt{(x-1)(x-2)}}$

Concept: U Substitution — we rewrite the quadratic inside the square root into a perfect square form to match a standard inverse hyperbolic or trigonometric integral.

First, expand the product:

Complete the square:

So the integral becomes:

Let $u = x - \frac{3}{2}$, so $du = dx$. The integral is now:

$\int \frac{du}{\sqrt{u^2 - \left(\frac{1}{2}\right)^2}}.$ …

Why This Approach Works

When you see a square root of a quadratic in the denominator, your first instinct should be: can I complete the square? The expression $(x-1)(x-2)$ expands to $x^2 - 3x + 2$, which is a quadratic. The square root of a quadratic often hides a standard inverse hyperbolic or trigonometric form.

Specifically, the integral $\int \frac{dx}{\sqrt{x^2 - a^2}}$ is a known result: it equals $\cosh^{-1}(x/a) + C$ (or $\log|x + \sqrt{x^2 - a^2}| + C$, if you prefer the logarithmic form). Our job is to manipulate the given integrand until it matches that pattern.

The trick is to complete the square inside the root, then use a substitution that turns the expression into something of the form $\sqrt{u^2 - a^2}$.

Step-by-Step Solution

1. Expand and complete the square

First, expand the product:

Now complete the square for $x^2 - 3x$. Half of $-3$ is $-\frac{3}{2}$, so we add and subtract $\left(\frac{3}{2}\right)^2 = \frac{9}{4}$:

So the integral becomes:

2. Choose a substitution to simplify

Let $u = x - \frac{3}{2}$. Then $du = dx$, and the integral is:

This is exactly the form $\int \frac{du}{\sqrt{u^2 - a^2}}$ with $a = \frac{1}{2}$.

3. Apply the formula …

Method: Integrate $1/\sqrt{\text{quadratic}}$ by completing the square

Any integral of the form $\displaystyle\int \frac{dx}{\sqrt{ax^2+bx+c}}$ is turned into one of three standard forms by completing the square inside the root.

Steps

Step 1: Make the $x^2$-coefficient $1$ and complete the square.

Write the radicand as a perfect square plus or minus a constant. If the quadratic is a product like $(x-p)(x-q)$, expand it first, then complete the square to reach $(x+h)^2 \pm k^2$.

Step 2: Substitute $u = x+h$ (so $du = dx$).

The integral becomes $\displaystyle\int \frac{du}{\sqrt{u^2 \pm k^2}}$ or $\displaystyle\int \frac{du}{\sqrt{k^2 - u^2}}$.

Step 3: Read off the correct standard form from the sign. …

Common Mistakes

Mistake 1: Trying to integrate before completing the square.

Why it's wrong: $\int dx/\sqrt{(x-p)(x-q)}$ has no direct antiderivative in product form. Correct approach: always expand to $ax^2+bx+c$ and complete the square first so it matches a standard $u^2\pm a^2$ form.

Mistake 2: Getting the wrong $a$ in the standard form.

Why it's wrong: after completing the square you get $u^2 - \left(\tfrac12\right)^2$, so $a=\tfrac12$, not $\tfrac14$. Correct approach: the subtracted constant is $a^2$, so $a$ is its square root. …