If `int(dx)/(x+x^(2011))=f(x)+C_(1) and int(x^(2009))/(1+x^(2010))dx=g(x)+C_(2)` (where `C_(1) and C_(2)` are constants of integration). Let h(x)=f(x)+g(x). If h(1)=0 then h(e) is equal to :

Answer

To solve the problem, we need to find the functions \( f(x) \) and \( g(x) \) from the given integrals, then combine them to find \( h(x) = f(x) + g(x) \), and finally evaluate \( h(e) \).

### Step 1: Find \( f(x) \)

We start with the integral:

\[ f(x) = \int \frac{dx}{x + x^{2011}} \]

Factor out \( x^{2011} \):

\[ = \int \frac{dx}{x(1 + \frac{1}{x^{2010}})} = \int \frac{dx}{x^{2011}(1 + \frac{1}{x^{2010}})} \]

Let \( y = \frac{1}{x^{2010}} + 1 \). Then, differentiate:

\[ dy = -\frac{2010}{x^{2011}} dx \implies dx = -\frac{1}{2010} x^{2011} dy \]

Substituting this into the integral gives:

\[ f(x) = \int -\frac{1}{2010} \frac{1}{y} dy = -\frac{1}{2010} \ln |y| + C_1 \]

Substituting back for \( y \):

\[ f(x) = -\frac{1}{2010} \ln \left(1 + \frac{1}{x^{2010}}\right) + C_1 \]

### Step 2: Find \( g(x) \)

Now we evaluate the second integral:

\[ g(x) = \int \frac{x^{2009}}{1 + x^{2010}} dx \]

Let \( y = 1 + x^{2010} \). Then, differentiate:

\[ dy = 2010 x^{2009} dx \implies dx = \frac{1}{2010 x^{2009}} dy \]

Substituting this into the integral gives:

\[ g(x) = \int \frac{1}{2010} \frac{1}{y} dy = \frac{1}{2010} \ln |y| + C_2 \]

Substituting back for \( y \):

\[ g(x) = \frac{1}{2010} \ln(1 + x^{2010}) + C_2 \]

### Step 3: Combine \( f(x) \) and \( g(x) \)

Now we combine \( f(x) \) and \( g(x) \):

\[ h(x) = f(x) + g(x) = -\frac{1}{2010} \ln \left(1 + \frac{1}{x^{2010}}\right) + \frac{1}{2010} \ln(1 + x^{2010}) + C \]

Where \( C = C_1 + C_2 \).

### Step 4: Simplify \( h(x) \)

Using properties of logarithms:

\[ h(x) = \frac{1}{2010} \left( \ln(1 + x^{2010}) - \ln \left(1 + \frac{1}{x^{2010}}\right) \right) + C \]

This can be simplified to:

\[ h(x) = \frac{1}{2010} \ln \left( \frac{(1 + x^{2010})}{\left(1 + \frac{1}{x^{2010}}\right)} \right) + C \]

### Step 5: Evaluate \( h(1) \)

Given \( h(1) = 0 \):

\[ h(1) = \frac{1}{2010} \ln \left( \frac{(1 + 1^{2010})}{\left(1 + \frac{1}{1^{2010}}\right)} \right) + C = 0 \]

This simplifies to:

\[ \frac{1}{2010} \ln(2) + C = 0 \implies C = -\frac{1}{2010} \ln(2) \]

### Step 6: Evaluate \( h(e) \)

Now we evaluate \( h(e) \):

\[ h(e) = \frac{1}{2010} \ln \left( \frac{(1 + e^{2010})}{\left(1 + \frac{1}{e^{2010}}\right)} \right) - \frac{1}{2010} \ln(2) \]

This simplifies to:

\[ h(e) = \frac{1}{2010} \left( \ln(1 + e^{2010}) - \ln(2) \right) \]

Using properties of logarithms:

\[ h(e) = \frac{1}{2010} \ln \left( \frac{1 + e^{2010}}{2} \right) \]

### Final Result

Thus, the value of \( h(e) \) is:

\[ h(e) = \frac{1}{2010} \ln \left( \frac{1 + e^{2010}}{2} \right) \]