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 given integral and find the value of \( k \), we start with the expression:

\[ \int \frac{x^2 - 1}{(x^4 + 3x^2 + 1) \tan^{-1}\left(\frac{x^2 + 1}{x}\right)} \, dx = k \log\left|\frac{\tan^{-1}(x^2 + 1)}{x}\right| + c \]

### Step 1: Simplify the Integral
We can simplify the integral by factoring out \( x^2 \) from the numerator and the denominator.

\[ \int \frac{x^2(1 - \frac{1}{x^2})}{x^2(x^2 + 3 + \frac{1}{x^2}) \tan^{-1}\left(\frac{x^2 + 1}{x}\right)} \, dx \]

This simplifies to:

\[ \int \frac{1 - \frac{1}{x^2}}{x^2 + 3 + \frac{1}{x^2}} \cdot \frac{1}{\tan^{-1}\left(\frac{x^2 + 1}{x}\right)} \, dx \]

### Step 2: Rewrite the Denominator
Next, we can rewrite the denominator:

\[ x^2 + 3 + \frac{1}{x^2} = \left(x + \frac{1}{x}\right)^2 + 2 \]

### Step 3: Use Substitution
Let:

\[ t = \tan^{-1}\left(\frac{x^2 + 1}{x}\right) \]

Differentiating \( t \):

\[ \frac{dt}{dx} = \frac{1}{1 + \left(\frac{x^2 + 1}{x}\right)^2} \cdot \left(\frac{d}{dx}\left(\frac{x^2 + 1}{x}\right)\right) \]

Calculating \( \frac{d}{dx}\left(\frac{x^2 + 1}{x}\right) \):

\[ \frac{d}{dx}\left(\frac{x^2 + 1}{x}\right) = \frac{(2x)(x) - (x^2 + 1)(1)}{x^2} = \frac{x^2 - 1}{x^2} \]

Thus,

\[ \frac{dt}{dx} = \frac{x^2 - 1}{x^2(1 + \left(\frac{x^2 + 1}{x}\right)^2)} \]

### Step 4: Substitute Back into the Integral
Substituting \( dt \) into the integral gives us:

\[ \int \frac{dt}{t} \]

### Step 5: Integrate
The integral of \( \frac{1}{t} \) is:

\[ \log|t| + C \]

Substituting back for \( t \):

\[ \log\left|\tan^{-1}\left(\frac{x^2 + 1}{x}\right)\right| + C \]

### Step 6: Compare with the Given Expression
Now, we compare:

\[ \log\left|\tan^{-1}\left(\frac{x^2 + 1}{x}\right)\right| + C = k \log\left|\frac{\tan^{-1}(x^2 + 1)}{x}\right| + c \]

From the comparison, we can see that:

\[ k = 1 \]

### Final Answer
Thus, the value of \( k \) is:

\[ \boxed{1} \]