`int_(0)^(1)(dx)/([ax+(1-x)b]^(2))=`

Options

A. ab
B. `(a)/(b)`
C. `(b)/(a)`
D. `(1)/(ab)`

Answer

To evaluate the integral

\[ I = \int_{0}^{1} \frac{dx}{(ax + (1-x)b)^2} \]

we will follow these steps:

### Step 1: Rewrite the integrand
First, we rewrite the expression in the denominator:

\[ ax + (1-x)b = ax + b - bx = (a-b)x + b \]

Thus, we can express the integral as:

\[ I = \int_{0}^{1} \frac{dx}{((a-b)x + b)^2} \]

### Step 2: Substitution
Next, we will use the substitution:

\[ t = (a-b)x + b \]

Now we differentiate \(t\) with respect to \(x\):

\[ \frac{dt}{dx} = a - b \quad \Rightarrow \quad dx = \frac{dt}{a-b} \]

### Step 3: Change the limits of integration
We need to change the limits of integration according to our substitution:

- When \(x = 0\):
\[ t = (a-b) \cdot 0 + b = b \]

- When \(x = 1\):
\[ t = (a-b) \cdot 1 + b = a \]

So, the new limits of integration are from \(b\) to \(a\).

### Step 4: Substitute into the integral
Now we substitute \(t\) and \(dx\) into the integral:

\[ I = \int_{b}^{a} \frac{1}{t^2} \cdot \frac{dt}{a-b} \]

This simplifies to:

\[ I = \frac{1}{a-b} \int_{b}^{a} \frac{dt}{t^2} \]

### Step 5: Evaluate the integral
The integral \(\int \frac{dt}{t^2}\) is:

\[ \int \frac{dt}{t^2} = -\frac{1}{t} \]

Now we evaluate it from \(b\) to \(a\):

\[ \int_{b}^{a} \frac{dt}{t^2} = \left[-\frac{1}{t}\right]_{b}^{a} = -\frac{1}{a} + \frac{1}{b} = \frac{1}{b} - \frac{1}{a} \]

### Step 6: Substitute back into the expression for \(I\)
Now substituting this back into our expression for \(I\):

\[ I = \frac{1}{a-b} \left(\frac{1}{b} - \frac{1}{a}\right) \]

### Step 7: Simplify the expression
We can simplify this further:

\[ I = \frac{1}{a-b} \cdot \frac{a-b}{ab} = \frac{1}{ab} \]

### Final Result
Thus, the value of the integral is:

\[ I = \frac{1}{ab} \]

---