If `y = int_(0)^(x^(2))ln(1+t)`, then find `(d^(2)y)/(dx^(2))`

Answer

To solve the problem, we need to find the second derivative of the function defined by the integral:

\[ y = \int_{0}^{x^2} \ln(1+t) \, dt \]

### Step 1: Differentiate \( y \) with respect to \( x \)

Using the Fundamental Theorem of Calculus and the chain rule, we differentiate \( y \):

\[ \frac{dy}{dx} = \frac{d}{dx} \left( \int_{0}^{x^2} \ln(1+t) \, dt \right) \]

According to the theorem, we have:

\[ \frac{dy}{dx} = \ln(1 + x^2) \cdot \frac{d}{dx}(x^2) - \ln(1 + 0) \cdot \frac{d}{dx}(0) \]

Since \( \ln(1 + 0) = 0 \) and \( \frac{d}{dx}(0) = 0 \), the second term vanishes. Thus, we have:

\[ \frac{dy}{dx} = \ln(1 + x^2) \cdot 2x \]

### Step 2: Differentiate \( \frac{dy}{dx} \) to find \( \frac{d^2y}{dx^2} \)

Now we need to differentiate \( \frac{dy}{dx} \):

\[ \frac{d^2y}{dx^2} = \frac{d}{dx} \left( 2x \ln(1 + x^2) \right) \]

Using the product rule, we have:

\[ \frac{d^2y}{dx^2} = 2 \ln(1 + x^2) + 2x \cdot \frac{d}{dx}(\ln(1 + x^2)) \]

Now we need to differentiate \( \ln(1 + x^2) \):

\[ \frac{d}{dx}(\ln(1 + x^2)) = \frac{1}{1 + x^2} \cdot \frac{d}{dx}(1 + x^2) = \frac{2x}{1 + x^2} \]

Substituting this back into our expression for \( \frac{d^2y}{dx^2} \):

\[ \frac{d^2y}{dx^2} = 2 \ln(1 + x^2) + 2x \cdot \frac{2x}{1 + x^2} \]

This simplifies to:

\[ \frac{d^2y}{dx^2} = 2 \ln(1 + x^2) + \frac{4x^2}{1 + x^2} \]

### Final Result

Thus, the second derivative is:

\[ \frac{d^2y}{dx^2} = 2 \ln(1 + x^2) + \frac{4x^2}{1 + x^2} \]