If `int(1)/((x^(2)+1)(x^(2)+4))dx=Atan^(-1)x+B" tan"^(-1)(x)/(2)+C` , then

Options

A. `A=1//3,B=-2//3`
B. `A=-1//3,B=2//3`
C. `A=-1//3,B=1//3`
D. `A=1//3,B=-1//6`

Answer

To solve the integral \(\int \frac{1}{(x^2 + 1)(x^2 + 4)} \, dx\), we will use the method of partial fractions.

### Step-by-Step Solution:

1. **Set up the Partial Fraction Decomposition**:
We express \(\frac{1}{(x^2 + 1)(x^2 + 4)}\) as:
\[ \frac{A}{x^2 + 1} + \frac{B}{x^2 + 4} \]
where \(A\) and \(B\) are constants to be determined.

2. **Combine the Right-Hand Side**:
To combine the fractions, we find a common denominator:
\[ \frac{A(x^2 + 4) + B(x^2 + 1)}{(x^2 + 1)(x^2 + 4)} = \frac{1}{(x^2 + 1)(x^2 + 4)} \]
This gives us the equation:
\[ A(x^2 + 4) + B(x^2 + 1) = 1 \]

3. **Expand and Collect Like Terms**:
Expanding the left side:
\[ Ax^2 + 4A + Bx^2 + B = (A + B)x^2 + (4A + B) \]
We set this equal to 1:
\[ (A + B)x^2 + (4A + B) = 1 \]

4. **Set Up the System of Equations**:
From the equation above, we can equate coefficients:
- For \(x^2\): \(A + B = 0\)
- For the constant term: \(4A + B = 1\)

5. **Solve the System of Equations**:
From \(A + B = 0\), we can express \(B\) as:
\[ B = -A \]
Substituting \(B = -A\) into the second equation:
\[ 4A - A = 1 \implies 3A = 1 \implies A = \frac{1}{3} \]
Then substituting back to find \(B\):
\[ B = -\frac{1}{3} \]

6. **Rewrite the Integral**:
Now we can rewrite the integral:
\[ \int \frac{1}{(x^2 + 1)(x^2 + 4)} \, dx = \int \left(\frac{1/3}{x^2 + 1} - \frac{1/3}{x^2 + 4}\right) \, dx \]

7. **Integrate Each Term**:
The integrals can be computed as follows:
\[ \int \frac{1/3}{x^2 + 1} \, dx = \frac{1}{3} \tan^{-1}(x) \]
\[ \int \frac{1/3}{x^2 + 4} \, dx = \frac{1}{3} \cdot \frac{1}{2} \tan^{-1}\left(\frac{x}{2}\right) = \frac{1}{6} \tan^{-1}\left(\frac{x}{2}\right) \]

8. **Combine the Results**:
Therefore, the integral becomes:
\[ \int \frac{1}{(x^2 + 1)(x^2 + 4)} \, dx = \frac{1}{3} \tan^{-1}(x) - \frac{1}{6} \tan^{-1}\left(\frac{x}{2}\right) + C \]

### Final Result:
Comparing with the given form \(A \tan^{-1}(x) + B \tan^{-1}\left(\frac{x}{2}\right) + C\), we find:
- \(A = \frac{1}{3}\)
- \(B = -\frac{1}{6}\)