`f(x)=int(5x^8+7x^6)/(x^2+1+2x^7)^2dx , if f(0)=0` then find `f(1)=1/k.` Then k is

Answer

To solve the problem, we need to evaluate the integral

\[ f(x) = \int \frac{5x^8 + 7x^6}{(x^2 + 1 + 2x^7)^2} \, dx \]

with the condition that \( f(0) = 0 \) and find \( f(1) = \frac{1}{k} \), then determine the value of \( k \).

### Step 1: Simplify the Integral
We start by simplifying the integrand. Notice that we can factor out \( x^7 \) from the denominator:

\[ x^2 + 1 + 2x^7 = x^7 \left( \frac{1}{x^5} + \frac{1}{x^7} + 2 \right) \]

Thus, the integral becomes:

\[ f(x) = \int \frac{5x^8 + 7x^6}{x^{14} \left( \frac{1}{x^5} + \frac{1}{x^7} + 2 \right)^2} \, dx \]

### Step 2: Change of Variables
Let \( t = \frac{1}{x^5} + \frac{1}{x^7} + 2 \). We differentiate \( t \):

\[ dt = \left( -\frac{5}{x^6} - \frac{7}{x^8} \right) dx \]

This gives us:

\[ dx = -\frac{x^6 x^8}{5x^6 + 7x^8} dt = -\frac{x^2}{5 + 7/x^2} dt \]

### Step 3: Rewrite the Integral
Now substituting back into the integral, we have:

\[ f(x) = -\int \frac{1}{t^2} dt \]

### Step 4: Integrate
The integral of \( -\frac{1}{t^2} \) is:

\[ f(x) = \frac{1}{t} + C \]

Substituting back for \( t \):

\[ f(x) = \frac{1}{\frac{1}{x^5} + \frac{1}{x^7} + 2} + C \]

### Step 5: Evaluate \( f(0) \)
To find \( C \), we use the condition \( f(0) = 0 \). As \( x \to 0 \):

\[ \frac{1}{\frac{1}{0} + \frac{1}{0} + 2} \to \frac{1}{2} \]

Thus, we have:

\[ 0 = \frac{1}{2} + C \implies C = -\frac{1}{2} \]

### Step 6: Final Expression for \( f(x) \)
Now we can express \( f(x) \):

\[ f(x) = \frac{1}{\frac{1}{x^5} + \frac{1}{x^7} + 2} - \frac{1}{2} \]

### Step 7: Evaluate \( f(1) \)
Now we evaluate \( f(1) \):

\[ f(1) = \frac{1}{\frac{1}{1^5} + \frac{1}{1^7} + 2} - \frac{1}{2} = \frac{1}{1 + 1 + 2} - \frac{1}{2} = \frac{1}{4} - \frac{1}{2} = \frac{1}{4} - \frac{2}{4} = -\frac{1}{4} \]

### Step 8: Find \( k \)
Given that \( f(1) = \frac{1}{k} \):

\[ -\frac{1}{4} = \frac{1}{k} \implies k = -4 \]

Thus, the value of \( k \) is:

\[ \boxed{4} \]