`lim_(n to oo)(n^(2)/((n^(2)+1)(n+1))+(n^(2))/((n^(2)+4)(n+2))+(n^(2))/((n^(2)+9)(n+3))+....+(n^(2))/((n^(2)+n^(2))(n+n)))` is equal to

Options

A. `pi/8 +1/4 log_(e )2`
B. `pi/4 +1/8 log_(e )2`
C. `pi/4 -1/8 log_(2)2`
D. `pi/8 +log_(e)sqrt(2)`

Answer

To solve the limit

\[ \lim_{n \to \infty} \left( \frac{n^2}{(n^2+1)(n+1)} + \frac{n^2}{(n^2+4)(n+2)} + \frac{n^2}{(n^2+9)(n+3)} + \ldots + \frac{n^2}{(n^2+n^2)(n+n)} \right), \]

we can express this limit as a summation:

\[ \lim_{n \to \infty} \sum_{r=1}^{n} \frac{n^2}{(n^2 + r^2)(n + r)}. \]

### Step 1: Factor out \( n^2 \)

We can rewrite each term in the summation by factoring out \( n^2 \) from both the numerator and the denominator:

\[ \frac{n^2}{(n^2 + r^2)(n + r)} = \frac{1}{\left(1 + \frac{r^2}{n^2}\right)\left(1 + \frac{r}{n}\right)}. \]

### Step 2: Rewrite the limit

Now, we can express the limit as:

\[ \lim_{n \to \infty} \frac{1}{n} \sum_{r=1}^{n} \frac{1}{\left(1 + \frac{r^2}{n^2}\right)\left(1 + \frac{r}{n}\right)}. \]

### Step 3: Recognize the Riemann sum

As \( n \to \infty \), the expression inside the summation resembles a Riemann sum for the integral of a function over the interval \([0, 1]\). We can let \( x = \frac{r}{n} \), which implies \( r = nx \) and \( \frac{dr}{n} = dx \):

\[ \lim_{n \to \infty} \sum_{r=1}^{n} \frac{1}{n} \frac{1}{\left(1 + x^2\right)\left(1 + x\right)}. \]

### Step 4: Set up the integral

Thus, we can express the limit as an integral:

\[ \int_{0}^{1} \frac{1}{(1 + x^2)(1 + x)} \, dx. \]

### Step 5: Simplify the integrand

We can simplify the integrand:

\[ \frac{1}{(1 + x^2)(1 + x)} = \frac{A}{1 + x} + \frac{B}{1 + x^2}. \]

Multiplying through by the denominator \( (1 + x)(1 + x^2) \) and equating coefficients, we can find \( A \) and \( B \).

### Step 6: Solve for \( A \) and \( B \)

After solving, we find:

\[ \int_{0}^{1} \left( \frac{A}{1 + x} + \frac{B}{1 + x^2} \right) \, dx. \]

### Step 7: Evaluate the integral

Evaluating the integrals separately:

1. For \( \int \frac{1}{1+x} \, dx \), we get \( \log(1+x) \).
2. For \( \int \frac{1}{1+x^2} \, dx \), we get \( \tan^{-1}(x) \).

### Step 8: Apply limits

Evaluating from 0 to 1 gives us:

\[ \left[ \log(2) - \log(1) \right] + \left[ \frac{\pi}{4} - 0 \right]. \]

### Final Result

Combining these results, we find:

\[ \frac{\pi}{4} + \frac{1}{4} \log(2). \]

Thus, the limit evaluates to:

\[ \frac{\pi}{8} + \frac{1}{4} \log(2). \]