`int(dx)/(7+6x-x^2)=`

Options

A. `(1)/(4)log|(7-x)/(1+x)|+c`
B. `(1)/(4)log|(1+x)/(7-x)|+c`
C. `(1)/(8)log|(7-x)/(1+x)|+c`
D. `(1)/(8)log|(1+x)/(7-x)|+c`

Answer

To solve the integral \(\int \frac{dx}{7 + 6x - x^2}\), we will follow these steps:

### Step 1: Rewrite the Denominator
First, we rewrite the quadratic expression in the denominator. We can rearrange it as:
\[ -x^2 + 6x + 7 = -(x^2 - 6x - 7) \]
Now, we will factor the quadratic expression \(x^2 - 6x - 7\).

### Step 2: Completing the Square
To complete the square for the expression \(x^2 - 6x - 7\), we take the coefficient of \(x\) (which is -6), halve it to get -3, and then square it to get 9. We can rewrite the expression as:
\[ x^2 - 6x + 9 - 9 - 7 = (x - 3)^2 - 16 \]
Thus, we have:
\[ 7 + 6x - x^2 = -((x - 3)^2 - 16) = 16 - (x - 3)^2 \]

### Step 3: Substitute into the Integral
Now we substitute this back into the integral:
\[ \int \frac{dx}{7 + 6x - x^2} = \int \frac{dx}{16 - (x - 3)^2} \]

### Step 4: Use the Standard Integral Formula
The integral \(\int \frac{dx}{a^2 - u^2}\) can be solved using the formula:
\[ \int \frac{dx}{a^2 - u^2} = \frac{1}{2a} \log\left|\frac{a + u}{a - u}\right| + C \]
In our case, \(a = 4\) and \(u = x - 3\). Therefore:
\[ \int \frac{dx}{16 - (x - 3)^2} = \frac{1}{2 \cdot 4} \log\left|\frac{4 + (x - 3)}{4 - (x - 3)}\right| + C \]
This simplifies to:
\[ \frac{1}{8} \log\left|\frac{x + 1}{7 - x}\right| + C \]

### Final Answer
Thus, the final answer to the integral is:
\[ \int \frac{dx}{7 + 6x - x^2} = \frac{1}{8} \log\left|\frac{x + 1}{7 - x}\right| + C \]

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