Let `int _( 1) ^(sqrt5)(x ^(2x ^(2)+1) +ln "("x ^(2x ^(2x ^(2+1)))")")dx =N.` Find the valur of `(N-6).`

Answer

To solve the problem, we need to evaluate the integral

\[ N = \int_{1}^{\sqrt{5}} \left( x^{2x^2 + 1} + \ln\left(x^{2x^{2x^2 + 1}}\right) \right) dx. \]

### Step 1: Simplifying the logarithmic term

Using the property of logarithms, we can simplify the logarithmic term:

\[ \ln\left(x^{2x^{2x^2 + 1}}\right) = 2x^{2x^2 + 1} \ln x. \]

Thus, we can rewrite the integral as:

\[ N = \int_{1}^{\sqrt{5}} \left( x^{2x^2 + 1} + 2x^{2x^2 + 1} \ln x \right) dx. \]

### Step 2: Factoring out the common term

We can factor out \(x^{2x^2 + 1}\):

\[ N = \int_{1}^{\sqrt{5}} x^{2x^2 + 1} \left( 1 + 2 \ln x \right) dx. \]

### Step 3: Substituting \(t = x^{x^2}\)

Let \(t = x^{x^2}\). Then, taking the natural logarithm of both sides gives:

\[ \ln t = x^2 \ln x. \]

Differentiating both sides with respect to \(x\):

\[ \frac{1}{t} \frac{dt}{dx} = 2x \ln x + x. \]

Thus, we have:

\[ dt = t(2x \ln x + x) dx. \]

### Step 4: Changing limits of integration

When \(x = 1\):

\[ t = 1^{1^2} = 1. \]

When \(x = \sqrt{5}\):

\[ t = (\sqrt{5})^{(\sqrt{5})^2} = 5^{\frac{5}{2}} = 5^{5/2}. \]

### Step 5: Expressing \(dx\) in terms of \(dt\)

From the expression for \(dt\):

\[ dx = \frac{dt}{t(2x \ln x + x)}. \]

### Step 6: Substituting back into the integral

Now, substituting back into the integral, we have:

\[ N = \int_{1}^{5^{5/2}} t \cdot \frac{1 + 2 \ln x}{2x \ln x + x} dt. \]

### Step 7: Evaluating the integral

The integral simplifies to:

\[ N = \int_{1}^{5^{5/2}} t dt = \left[\frac{t^2}{2}\right]_{1}^{5^{5/2}} = \frac{(5^{5/2})^2}{2} - \frac{1^2}{2} = \frac{25^5}{2} - \frac{1}{2}. \]

Calculating \(N\):

\[ N = \frac{3125 - 1}{2} = \frac{3124}{2} = 1562. \]

### Step 8: Finding \(N - 6\)

Finally, we find:

\[ N - 6 = 1562 - 6 = 1556. \]

### Final Answer

\[ \boxed{1556}. \]