`int_(0)^(1)(1)/(x^(2)+4x+5)dx =`

Options

A. `tan^(-1)((1)/(7))`
B. `log((1)/(7))`
C. `sin^(-1)((1)/(7))`
D. log 7

Answer

To solve the integral \( \int_{0}^{1} \frac{1}{x^2 + 4x + 5} \, dx \), we will follow these steps:

### Step 1: Rewrite the quadratic expression
We start with the expression in the denominator:
\[ x^2 + 4x + 5 \]
We can complete the square for the quadratic:
\[ x^2 + 4x + 5 = (x^2 + 4x + 4) + 1 = (x + 2)^2 + 1 \]

### Step 2: Substitute the expression into the integral
Now we can rewrite the integral:
\[ \int_{0}^{1} \frac{1}{(x + 2)^2 + 1} \, dx \]

### Step 3: Make a substitution
Let \( t = x + 2 \). Then, \( dx = dt \). We also need to change the limits of integration:
- When \( x = 0 \), \( t = 0 + 2 = 2 \)
- When \( x = 1 \), \( t = 1 + 2 = 3 \)

So, the integral becomes:
\[ \int_{2}^{3} \frac{1}{t^2 + 1} \, dt \]

### Step 4: Integrate the new expression
The integral \( \int \frac{1}{t^2 + 1} \, dt \) is a standard integral, which equals \( \tan^{-1}(t) \). Thus, we have:
\[ \int_{2}^{3} \frac{1}{t^2 + 1} \, dt = \tan^{-1}(t) \bigg|_{2}^{3} \]

### Step 5: Evaluate the definite integral
Now we evaluate the definite integral:
\[ \tan^{-1}(3) - \tan^{-1}(2) \]

### Step 6: Use the formula for the difference of arctangents
To simplify \( \tan^{-1}(3) - \tan^{-1}(2) \), we can use the formula:
\[ \tan^{-1}(a) - \tan^{-1}(b) = \tan^{-1}\left(\frac{a - b}{1 + ab}\right) \]
Here, \( a = 3 \) and \( b = 2 \):
\[ \tan^{-1}(3) - \tan^{-1}(2) = \tan^{-1}\left(\frac{3 - 2}{1 + 3 \cdot 2}\right) = \tan^{-1}\left(\frac{1}{1 + 6}\right) = \tan^{-1}\left(\frac{1}{7}\right) \]

### Final Answer
Thus, the value of the integral is:
\[ \int_{0}^{1} \frac{1}{x^2 + 4x + 5} \, dx = \tan^{-1}\left(\frac{1}{7}\right) \]

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