If `int(1-x^(7))/(x(1+x^(7)))dx=alog_(e)|x|+blog_(e)|x^(7)+1|+c,` then

Options

A. `a=1`
B. `a= -1`
C. `b=(2)/(7)`
D. `b= -(2)/(7)`

Explanation

To solve the integral

\[ \int \frac{1 - x^7}{x(1 + x^7)} \, dx, \]

we can start by simplifying the integrand.

### Step 1: Simplify the integrand

We can rewrite the integrand as follows:

\[ \frac{1 - x^7}{x(1 + x^7)} = \frac{1}{x(1 + x^7)} - \frac{x^7}{x(1 + x^7)} = \frac{1}{x(1 + x^7)} - \frac{x^6}{1 + x^7}. \]

So we can express the integral as:

\[ \int \left( \frac{1}{x(1 + x^7)} - \frac{x^6}{1 + x^7} \right) \, dx. \]

### Step 2: Split the integral

Now we can split the integral into two parts:

\[ \int \frac{1}{x(1 + x^7)} \, dx - \int \frac{x^6}{1 + x^7} \, dx. \]

### Step 3: Solve the first integral

For the first integral, we can use the substitution \( u = 1 + x^7 \), then \( du = 7x^6 \, dx \) or \( dx = \frac{du}{7x^6} \). We also have \( x^6 = (u - 1)^{6/7} \).

Thus, we can rewrite the first integral:

\[ \int \frac{1}{x(1 + x^7)} \, dx = \int \frac{1}{x u} \cdot \frac{du}{7x^6} = \frac{1}{7} \int \frac{1}{u} \, du = \frac{1}{7} \ln |u| + C_1 = \frac{1}{7} \ln |1 + x^7| + C_1. \]

### Step 4: Solve the second integral

For the second integral, we can use the same substitution \( u = 1 + x^7 \), which gives:

\[ \int \frac{x^6}{1 + x^7} \, dx = \int \frac{x^6}{u} \cdot \frac{du}{7x^6} = \frac{1}{7} \int \frac{1}{u} \, du = \frac{1}{7} \ln |u| + C_2 = \frac{1}{7} \ln |1 + x^7| + C_2. \]

### Step 5: Combine the results

Now we can combine the results of both integrals:

\[ \int \frac{1 - x^7}{x(1 + x^7)} \, dx = \frac{1}{7} \ln |x| - \frac{1}{7} \ln |1 + x^7| + C. \]

### Step 6: Final expression

This can be simplified to:

\[ \int \frac{1 - x^7}{x(1 + x^7)} \, dx = \frac{1}{7} \ln \left| \frac{x}{1 + x^7} \right| + C. \]

### Conclusion

Thus, we have:

\[ \int \frac{1 - x^7}{x(1 + x^7)} \, dx = a \ln |x| + b \ln |1 + x^7| + C, \]

where \( a = \frac{1}{7} \) and \( b = -\frac{1}{7} \).