Euler's substitution:
Integrals of the form `intR(x, sqrt(ax^(2)+bx+c))dx` are claculated with the aid of one of the following three Euler substitutions:
i. `sqrt(ax^(2)+bx+c)=t+-x sqrt(a)if a gt 0`
ii. `sqrt(ax^(2)+bx+c)=tx+-x sqrt(c)if c gt 0`
iii. `sqrt(ax^(2)+bx+c)=(x-a)t if ax^(2)+bx+c=a(x-a)(x-b)` i.e., if `alpha` is real root of `ax^(2)+bx+c=0`
`(xdx)/(sqrt(7x-10-x^(2))^3)`can be evaluated by substituting for x as

Options

A. x=( 5+ 2 t^2 )/(t^2 +1)
B. x=( 5 - t^2 )/(t^2 +2)
C. x=(2 t^2 - 5 )/(3t^2 -1)
D. none of these

Explanation

To solve the integral \(\int \frac{x \, dx}{\sqrt{7x - 10 - x^2}^3}\) using Euler's substitution, we will follow these steps:

### Step 1: Identify the form of the integral
We have the integral in the form \(\int \frac{x \, dx}{\sqrt{7x - 10 - x^2}^3}\). We need to simplify the expression under the square root.

### Step 2: Simplify the expression under the square root
We start with the expression \(7x - 10 - x^2\). We can rewrite it as:
\[ -x^2 + 7x - 10 = - (x^2 - 7x + 10) \]
Next, we can factor the quadratic:
\[ x^2 - 7x + 10 = (x - 2)(x - 5) \]
Thus, we have:
\[ 7x - 10 - x^2 = - (x - 2)(x - 5) \]

### Step 3: Apply Euler's substitution
Since we have a quadratic expression under the square root, we can use Euler's substitution. According to the third substitution, we set:
\[ \sqrt{7x - 10 - x^2} = (x - 2)t \]
This gives us:
\[ 7x - 10 - x^2 = -(x - 2)(x - 5) = -(x - 2)t^2 \]

### Step 4: Solve for \(x\)
From our substitution, we can express \(x\) in terms of \(t\):
\[ x - 2 = \frac{\sqrt{7x - 10 - x^2}}{t} \]
Rearranging gives:
\[ x = 2 + (x - 2)t \]
Solving for \(x\) yields:
\[ x = \frac{5 + 2t^2}{t^2 + 1} \]

### Step 5: Substitute back into the integral
Now we substitute \(x = \frac{5 + 2t^2}{t^2 + 1}\) back into the integral and simplify.

### Final Step: Evaluate the integral
After substitution, we can evaluate the integral using the new variable \(t\). The exact evaluation will depend on the new limits and the transformed differential \(dx\).