Q.Integrate the following function: $\frac{1}{\sqrt{7 - 6x - x^2}}$

The key idea is Integration by Completing the Square — rewriting the quadratic inside the square root to match the form $a^2 - (x + b)^2$, which integrates to $\sin^{-1}$.

First, rewrite the quadratic:

So the integral becomes:

$\int \frac{dx}{\sqrt{16 - (x+3)^2}}.$ …

When you see a quadratic inside a square root in the denominator, your first instinct should be: can I complete the square? The reason is simple. The standard integrals we know — like $\int \frac{dx}{\sqrt{a^2 - x^2}} = \arcsin(x/a) + C$ — are all built around perfect squares. A messy quadratic like $7 - 6x - x^2$ hides a perfect square inside it. Completing the square reveals that structure, letting you match the integral to a known form.

The trick is to handle the negative sign in front of $x^2$ carefully. Here, the quadratic is $-x^2 - 6x + 7$. Factor out the negative from the $x^2$ and $x$ terms, then complete the square inside the parentheses.

1. Rewrite the quadratic Start with $7 - 6x - x^2$. It’s easier to group the $x$ terms:

Now complete the square for $x^2 + 6x$. Half of 6 is 3, and $3^2 = 9$. So:

Substitute back:

2. Rewrite the integral The original integral becomes:

This is exactly the form $\int \frac{dx}{\sqrt{a^2 - u^2}}$ with $a = 4$ and $u = x+3$.

3. Apply the standard formula We know:

Here $du = dx$ (since $u = x+3$ gives $du = dx$), so:

$\int \frac{dx}{\sqrt{16 - (x+3)^2}} = \arcsin\left(\frac{x+3}{4}\right) + C$ …

Method: Complete the square with a negative $x^2$, then use the arcsine form

When the quadratic under the root has a $-x^2$ leading term, factor out the minus and complete the square to reach $a^2-(\text{linear})^2$, which integrates to an inverse sine.

Steps

Step 1: Rearrange and complete the square.

Step 2: Substitute the linear block.

With $u=x+3$, $du=dx$, the integral is $\displaystyle\int\frac{du}{\sqrt{16-u^2}}$.

Step 3: Apply the arcsine standard form. …

Common Mistakes

Mistake 1: Completing the square without first factoring out the $-1$ on $x^2$.

Why it's wrong: with a $-x^2$ term you must factor the minus before completing the square, or the constant lands with the wrong sign and you cannot reach $a^2-u^2$. Correct approach: $7-6x-x^2=-(x^2+6x-7)=16-(x+3)^2$.

Mistake 2: Using the $\log$/$\sinh^{-1}$ form instead of arcsine. …