Give step-by-step solution with explanation and final answer:dx
whats [————7
2+

Question

Give step-by-step solution with explanation and final answer:dx
whats [————7
2+

VIPSolutions · Accepted Answer

Let's solve the integral step by step:

$$
\int \frac{dx}{\sqrt{x^2 - 2x + 8}}
$$

### Step 1: Complete the Square

The expression under the square root is $x^2 - 2x + 8$.

Complete the square:
$$
x^2 - 2x + 8 = (x^2 - 2x + 1) + 7 = (x - 1)^2 + 7
$$

So the integral becomes:
$$
\int \frac{dx}{\sqrt{(x-1)^2 + 7}}
$$

### Step 2: Use a Substitution

Let $u = x - 1$, so $du = dx$.

Now the integral is:
$$
\int \frac{du}{\sqrt{u^2 + 7}}
$$

### Step 3: Recognize the Standard Form

This is a standard integral:
$$
\int \frac{du}{\sqrt{u^2 + a^2}} = \ln|u + \sqrt{u^2 + a^2}| + C
$$
where $a = \sqrt{7}$.

### Step 4: Substitute Back

So,
$$
\int \frac{du}{\sqrt{u^2 + 7}} = \ln|u + \sqrt{u^2 + 7}| + C
$$

Recall that $u = x - 1$:

$$
\int \frac{dx}{\sqrt{x^2 - 2x + 8}} = \ln|x - 1 + \sqrt{(x-1)^2 + 7}| + C
$$

---

### **Final Answer**

$$
\boxed{
\int \frac{dx}{\sqrt{x^2 - 2x + 8}} = \ln|x - 1 + \sqrt{x^2 - 2x + 8}| + C
}
$$

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