Q.$\int \sqrt{x^2-8x+7} dx$ is equal to (A) $\frac{1}{2}(x-4)\sqrt{x^2-8x+7} + 9\log |x-4+\sqrt{x^2-8x+7}| + C$ (B) $\frac{1}{2}(x-4)\sqrt{x^2-8x+7} + 9\log |x+4+\sqrt{x^2-8x+7}| + C$ (C) $\frac{1}{2}(x-4)\sqrt{x^2-8x+7} - 3\sqrt{2}\log |x-4+\sqrt{x^2-8x+7}| + C$ (D) $\frac{1}{2}(x-4)\sqrt{x^2-8x+7} - \frac{9}{2}\log |x-4+\sqrt{x^2-8x+7}| + C$

Concept: Integration by completing the square — rewrite the quadratic inside the square root as a perfect square minus a constant, then use the standard form $\int \sqrt{u^2 - a^2}\,du$.

Step 1: Complete the square inside the radicand:

So the integral becomes $\int \sqrt{(x-4)^2 - 3^2}\,dx$.

Step 2: Use the standard formula:

with $u = x-4$ and $a = 3$.

Step 3: Substitute back: …

When you see a quadratic inside a square root, your first instinct should be to complete the square. Why? Because the expression $\sqrt{x^2 - 8x + 7}$ doesn't match any standard integration formula directly. But if we rewrite it as $\sqrt{(x-4)^2 - 9}$, it becomes $\sqrt{u^2 - a^2}$ — a form with a known antiderivative.

The key formula we need is:

This formula comes from a trigonometric substitution ($u = a\sec\theta$), but you don't need to re-derive it every time — just apply it carefully.

Let's work through it step by step.

1. Complete the square inside the radical. $x^2 - 8x + 7 = (x^2 - 8x + 16) - 16 + 7 = (x-4)^2 - 9$ So the integral becomes:

2. Make a substitution to match the standard form. Let $u = x-4$, so $du = dx$. Then:

Here $a^2 = 9$, so $a = 3$.

3. Apply the standard formula. Using $\int \sqrt{u^2 - a^2}\,du = \frac{u}{2}\sqrt{u^2 - a^2} - \frac{a^2}{2}\log|u + \sqrt{u^2 - a^2}| + C$ with $a^2 = 9$:

4. Substitute back $u = x-4$. …

Method: Complete the square inside an MCQ radical, then match the template

For a multiple-choice $\int\sqrt{\text{quadratic}}\,dx$, complete the square to fix $a^{2}$, then compare the $\log$-coefficient against the options.

Steps

Step 1: Complete the square. $x^{2}-8x+7=(x-4)^{2}-9$, so $t=x-4$, $a^{2}=9$.

Step 2: Apply the $\sqrt{t^{2}-a^{2}}$ standard form.

Step 3: Substitute $t=x-4$, $a^{2}=9$. …

Common Mistakes

Mistake 1: Getting the sign of the log term wrong.

Why it's wrong: the $t^{2}-a^{2}$ form has a minus $\dfrac{a^{2}}{2}\log$; a plus sign gives distractor (A)/(B). Correct approach: keep the minus.

Mistake 2: Using $a=\sqrt9=3$ inside the log-coefficient instead of $a^{2}=9$. …