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

Answer

To solve the integral

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

we will break it down step by step.

### Step 1: Simplify the logarithmic term

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

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

Thus, we can rewrite \(N\) as:

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

### Step 2: Combine the terms

Now, we can combine the terms inside the integral:

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

### Step 3: Factor out common terms

Notice that \(x^{2x^2 + 1}\) can be factored out as follows:

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

### Step 4: Substitution

Let \(t = x^{x^2}\). Then, differentiating both sides gives:

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

This means we can express \(dx\) in terms of \(dt\):

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

### Step 5: Change the limits

When \(x = 1\), \(t = 1^{1^2} = 1\).
When \(x = \sqrt{5}\), \(t = (\sqrt{5})^{(\sqrt{5})^2} = 5^{\frac{5}{2}} = 5\).

### Step 6: Rewrite the integral

Now we can rewrite the integral in terms of \(t\):

\[ N = \int_{1}^{5} \frac{t^2}{t} dt = \int_{1}^{5} t dt. \]

### Step 7: Evaluate the integral

Now, we can evaluate the integral:

\[ N = \left[ \frac{t^2}{2} \right]_{1}^{5} = \frac{5^2}{2} - \frac{1^2}{2} = \frac{25}{2} - \frac{1}{2} = \frac{24}{2} = 12. \]

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

Finally, we find \(N - 6\):

\[ N - 6 = 12 - 6 = 6. \]

Thus, the value of \(N - 6\) is

\[ \boxed{6}. \]