Evaluate `int (dx)/((x-3)^(3)sqrt(x^(2)-6x+10))`.

Answer

To evaluate the integral

\[ I = \int \frac{dx}{(x-3)^3 \sqrt{x^2 - 6x + 10}}, \]

we will follow these steps:

### Step 1: Simplify the square root
First, we simplify the expression under the square root:

\[ x^2 - 6x + 10 = (x^2 - 6x + 9) + 1 = (x - 3)^2 + 1. \]

Thus, we can rewrite the integral as:

\[ I = \int \frac{dx}{(x-3)^3 \sqrt{(x-3)^2 + 1}}. \]

### Step 2: Substitution
Let \( t = x - 3 \), which implies \( dx = dt \). The integral becomes:

\[ I = \int \frac{dt}{t^3 \sqrt{t^2 + 1}}. \]

### Step 3: Rewrite the integral
Now we can express the integral in a more manageable form:

\[ I = \int \frac{dt}{t^3 \sqrt{t^2 + 1}}. \]

### Step 4: Further substitution
We can use the substitution \( u = \sqrt{t^2 + 1} \). Then, we have:

\[ u^2 = t^2 + 1 \implies t^2 = u^2 - 1 \implies t = \sqrt{u^2 - 1}. \]

Differentiating gives:

\[ dt = \frac{u}{\sqrt{u^2 - 1}} du. \]

### Step 5: Change the variable in the integral
Substituting \( t \) and \( dt \) into the integral:

\[ I = \int \frac{\frac{u}{\sqrt{u^2 - 1}} du}{(\sqrt{u^2 - 1})^3 u} = \int \frac{du}{(u^2 - 1)^{3/2}}. \]

### Step 6: Integrate
The integral can now be solved using the standard integral:

\[ \int \frac{du}{(u^2 - 1)^{3/2}} = -\frac{1}{\sqrt{u^2 - 1}} + C. \]

### Step 7: Back substitution
Now we substitute back \( u = \sqrt{(x - 3)^2 + 1} \):

\[ I = -\frac{1}{\sqrt{(x - 3)^2 + 1}} + C. \]

### Final Answer
Thus, the evaluated integral is:

\[ I = -\frac{1}{\sqrt{(x - 3)^2 + 1}} + C. \]