Let `int _(0)^(1) (4x ^(3) (1+(x ^(4)) ^(2010)))/((1+x^(4))^(2012))dx = (lamda)/(mu)`
where `lamda and mu` are relatively prime positive integers. Find unit digit of `mu`.

Answer

To solve the integral

\[ I = \int_0^1 \frac{4x^3(1 + (x^4)^{2010})}{(1 + x^4)^{2012}} \, dx \]

we will use a substitution method. Let's follow the steps:

### Step 1: Substitution
Let \( t = x^4 \). Then, differentiating both sides gives us:

\[ dt = 4x^3 \, dx \quad \Rightarrow \quad dx = \frac{dt}{4x^3} \]

Now, we also need to change the limits of integration. When \( x = 0 \), \( t = 0^4 = 0 \). When \( x = 1 \), \( t = 1^4 = 1 \).

### Step 2: Change of Variables
Substituting \( t \) into the integral, we have:

\[ I = \int_0^1 \frac{4x^3(1 + t^{2010})}{(1 + t)^{2012}} \cdot \frac{dt}{4x^3} \]

The \( 4x^3 \) terms cancel out:

\[ I = \int_0^1 \frac{1 + t^{2010}}{(1 + t)^{2012}} \, dt \]

### Step 3: Split the Integral
We can split the integral into two parts:

\[ I = \int_0^1 \frac{1}{(1 + t)^{2012}} \, dt + \int_0^1 \frac{t^{2010}}{(1 + t)^{2012}} \, dt \]

### Step 4: Evaluate the First Integral
The first integral can be evaluated using the formula for the beta function or directly:

\[ \int_0^1 (1 + t)^{-n} \, dt = \frac{1}{n-1} \text{ for } n > 1 \]

Here, \( n = 2012 \):

\[ \int_0^1 \frac{1}{(1 + t)^{2012}} \, dt = \frac{1}{2011} \]

### Step 5: Evaluate the Second Integral
For the second integral, we can use integration by parts or recognize it as a beta function:

\[ \int_0^1 \frac{t^{2010}}{(1 + t)^{2012}} \, dt = B(2011, 1) = \frac{\Gamma(2011)\Gamma(1)}{\Gamma(2012)} = \frac{2010!}{2011!} = \frac{1}{2011} \]

### Step 6: Combine the Results
Now, we can combine both integrals:

\[ I = \frac{1}{2011} + \frac{1}{2011} = \frac{2}{2011} \]

### Step 7: Express in Terms of Lambda and Mu
We have \( I = \frac{\lambda}{\mu} \) where \( \lambda = 2 \) and \( \mu = 2011 \). Since 2 and 2011 are relatively prime, we can conclude:

\[ \lambda = 2, \quad \mu = 2011 \]

### Step 8: Find the Unit Digit of Mu
The unit digit of \( \mu = 2011 \) is:

\[ \text{Unit digit of } 2011 = 1 \]

### Final Answer
Thus, the unit digit of \( \mu \) is:

\[ \boxed{1} \]