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. A= -1/3
C. B= 1/6
D. B= -1/6

Answer

To solve the integral \( \int \frac{1}{(x^2 + 1)(x^2 + 4)} \, dx \) and express it in the form \( A \tan^{-1}(x) + B \tan^{-1}\left(\frac{x}{2}\right) + C \), we will follow these steps:

### Step 1: Rewrite the Integral
We start with the integral:
\[ \int \frac{1}{(x^2 + 1)(x^2 + 4)} \, dx \]
To simplify the integration, we can multiply the numerator and denominator by 3:
\[ \int \frac{3}{3(x^2 + 1)(x^2 + 4)} \, dx = \frac{1}{3} \int \frac{3}{(x^2 + 1)(x^2 + 4)} \, dx \]

### Step 2: Split the Fraction
Next, we can express the numerator \( 3 \) as \( (x^2 + 4) - (x^2 + 1) \):
\[ \frac{3}{(x^2 + 1)(x^2 + 4)} = \frac{x^2 + 4}{(x^2 + 1)(x^2 + 4)} - \frac{x^2 + 1}{(x^2 + 1)(x^2 + 4)} \]
This allows us to split the integral:
\[ \frac{1}{3} \left( \int \frac{1}{x^2 + 1} \, dx - \int \frac{1}{x^2 + 4} \, dx \right) \]

### Step 3: Integrate Each Part
Now we can integrate each part separately:
1. The integral of \( \frac{1}{x^2 + 1} \) is \( \tan^{-1}(x) \).
2. The integral of \( \frac{1}{x^2 + 4} \) can be rewritten as \( \frac{1}{2} \cdot \frac{1}{\left(\frac{x}{2}\right)^2 + 1} \), which gives us \( \frac{1}{2} \tan^{-1}\left(\frac{x}{2}\right) \).

Putting it all together:
\[ \frac{1}{3} \left( \tan^{-1}(x) - \frac{1}{2} \tan^{-1}\left(\frac{x}{2}\right) \right) + C \]

### Step 4: Simplify the Expression
Distributing \( \frac{1}{3} \):
\[ \frac{1}{3} \tan^{-1}(x) - \frac{1}{6} \tan^{-1}\left(\frac{x}{2}\right) + C \]

### Step 5: Identify Constants A and B
From the expression, we can identify:
- \( A = \frac{1}{3} \)
- \( B = -\frac{1}{6} \)

### Final Answer
Thus, the values of \( A \) and \( B \) are:
- \( A = \frac{1}{3} \)
- \( B = -\frac{1}{6} \)