`int (dx)/(x ^(2014)+x)=1/p ln ((x ^(q))/(1+ x ^(r)))+C` where `p,q r in N` then the value of `(p+q+r)` equals (Where C is constant of integration)

Options

A. 6039
B. 6048
C. 6047
D. 6021

Answer

To solve the integral

\[ I = \int \frac{dx}{x^{2014} + x} \]

we will follow these steps:

### Step 1: Factor out \(x^{2014}\)
We can factor out \(x^{2014}\) from the denominator:

\[ I = \int \frac{dx}{x^{2014}(1 + \frac{1}{x^{2013}})} = \int \frac{1}{x^{2014}} \cdot \frac{dx}{1 + \frac{1}{x^{2013}}} \]

### Step 2: Substitute \(u = 1 + x^{-2013}\)
Let \(u = 1 + x^{-2013}\). Then, we differentiate \(u\) with respect to \(x\):

\[ \frac{du}{dx} = -2013 x^{-2014} \]

This implies:

\[ du = -2013 x^{-2014} dx \quad \Rightarrow \quad dx = -\frac{1}{2013} x^{2014} du \]

### Step 3: Substitute \(dx\) and simplify the integral
Now substituting \(dx\) in the integral:

\[ I = \int \frac{1}{x^{2014}} \cdot \frac{-\frac{1}{2013} x^{2014} du}{u} = -\frac{1}{2013} \int \frac{du}{u} \]

### Step 4: Integrate
The integral of \(\frac{1}{u}\) is \(\ln |u|\):

\[ I = -\frac{1}{2013} \ln |u| + C \]

### Step 5: Substitute back for \(u\)
Substituting back \(u = 1 + x^{-2013}\):

\[ I = -\frac{1}{2013} \ln |1 + x^{-2013}| + C \]

### Step 6: Rewrite the expression
We can rewrite the logarithm:

\[ I = \frac{1}{2013} \ln \left(\frac{1}{1 + x^{-2013}}\right) + C = \frac{1}{2013} \ln \left(\frac{x^{2013}}{1 + x^{2013}}\right) + C \]

### Step 7: Compare with the given form
The integral is given in the form:

\[ I = \frac{1}{p} \ln \left(\frac{x^q}{1 + x^r}\right) + C \]

From our expression, we can see:

- \(p = 2013\)
- \(q = 2013\)
- \(r = 2013\)

### Step 8: Calculate \(p + q + r\)
Now we can find:

\[ p + q + r = 2013 + 2013 + 2013 = 6039 \]

Thus, the final answer is:

\[ \boxed{6039} \]