Statement I If `a gt 0 ` and `b^(2)- 4ac lt 0`, then the value of the integral `int(dx)/(ax^(2)+bx+c)` will be of the type
`mu tan^(-1) . (x+A)/(B)+C`, where A, B, C, `mu` are constants.
Statement II If `a gt 0, b^(2)- 4ac lt 0`, then `ax^(2)+bx +C` can be written as sum of two squares .

Options

A. (a)Statement I is true, Statement II is also true , Statement II is the correct explanation of Statement I.
B. (b)Statement I is true, Statement II is also true, Statement II is not the correct explanation of Statement I.
C. (c)Statement I is true, Statement II is false.
D. (d)Statement I is false, Statement II is true .

Answer

To solve the problem, we need to analyze the statements provided and derive the necessary conclusions step by step.

### Step 1: Analyze Statement I
We need to evaluate the integral

\[ I = \int \frac{dx}{ax^2 + bx + c} \]

Given that \( a > 0 \) and \( b^2 - 4ac < 0 \), this implies that the quadratic \( ax^2 + bx + c \) does not have real roots and is always positive (since \( a > 0 \)).

### Step 2: Completing the Square
To simplify the integral, we can complete the square for the quadratic expression in the denominator:

\[ ax^2 + bx + c = a\left(x^2 + \frac{b}{a}x + \frac{c}{a}\right) \]

Completing the square inside the parentheses:

\[ x^2 + \frac{b}{a}x = \left(x + \frac{b}{2a}\right)^2 - \left(\frac{b^2}{4a^2} - \frac{c}{a}\right) \]

Thus, we can rewrite the quadratic as:

\[ ax^2 + bx + c = a\left(\left(x + \frac{b}{2a}\right)^2 + \left(\frac{4ac - b^2}{4a^2}\right)\right) \]

Let \( k^2 = \frac{4ac - b^2}{4a^2} \) (which is positive since \( b^2 - 4ac < 0 \)), we have:

\[ ax^2 + bx + c = a\left(\left(x + \frac{b}{2a}\right)^2 + k^2\right) \]

### Step 3: Substitute in the Integral
Now substituting this back into the integral:

\[ I = \int \frac{dx}{a\left(\left(x + \frac{b}{2a}\right)^2 + k^2\right)} = \frac{1}{a} \int \frac{dx}{\left(x + \frac{b}{2a}\right)^2 + k^2} \]

### Step 4: Use the Arctangent Integral Formula
The integral

\[ \int \frac{dx}{x^2 + k^2} = \frac{1}{k} \tan^{-1}\left(\frac{x}{k}\right) + C \]

Applying this formula, we get:

\[ I = \frac{1}{a} \cdot \frac{1}{k} \tan^{-1}\left(\frac{x + \frac{b}{2a}}{k}\right) + C \]

### Step 5: Conclusion for Statement I
Thus, the integral can be expressed in the form:

\[ I = \mu \tan^{-1}\left(\frac{x + A}{B}\right) + C \]

where \( \mu = \frac{1}{ak} \), \( A = \frac{b}{2a} \), \( B = k \), and \( C \) is a constant. Therefore, Statement I is correct.

### Step 6: Analyze Statement II
For Statement II, we have:

If \( a > 0 \) and \( b^2 - 4ac < 0 \), then \( ax^2 + bx + c \) can indeed be expressed as a sum of two squares. This follows from the completion of the square we performed earlier, which shows that the expression is always positive and can be represented as a sum of squares.

### Conclusion
Both statements are correct.