If `int((x^(2)-1)dx)/((x^(4)+3x^(2)+1)Tan^(-1)((x^(2)+1)/(x)))=klog|tan^(-1)""(x^(2)+1)/x|+c`, then k is equal to

Options

A. 1
B. 2
C. 3
D. 5

Answer

To solve the integral given in the question, we will follow these steps:

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

### Step 2: Substitute for Simplification
Let:
\[ t = \tan^{-1}\left(\frac{x^2 + 1}{x}\right) \]
Then, we differentiate \(t\) with respect to \(x\):
\[ dt = \frac{1}{1 + \left(\frac{x^2 + 1}{x}\right)^2} \cdot \left(\frac{d}{dx}\left(\frac{x^2 + 1}{x}\right)\right) \, dx \]
Calculating the derivative:
\[ \frac{d}{dx}\left(\frac{x^2 + 1}{x}\right) = \frac{(2x)(x) - (x^2 + 1)(1)}{x^2} = \frac{x^2 - 1}{x^2} \]
Thus, we have:
\[ dt = \frac{x^2 - 1}{x^2 + (x^2 + 1)^2} \, dx \]

### Step 3: Substitute Back into the Integral
Now, we can rewrite the integral in terms of \(t\):
\[ \int \frac{dt}{t} \]
This integral simplifies to:
\[ \ln |t| + C \]

### Step 4: Substitute \(t\) Back
Substituting back for \(t\):
\[ \ln \left| \tan^{-1}\left(\frac{x^2 + 1}{x}\right) \right| + C \]

### Step 5: Compare with Given Expression
We are given that:
\[ \int \frac{x^2 - 1}{x^4 + 3x^2 + 1} \, dx = k \ln \left| \tan^{-1}\left(\frac{x^2 + 1}{x}\right) \right| + C \]
From our calculation, we see that:
\[ k = 1 \]

### Final Answer
Thus, the value of \(k\) is:
\[ \boxed{1} \]