Let `I= int_(0)^(1) (e^(x))/( x+1)` dx, then the vlaue of the intergral ` int_(0)^(1) (xe^(x^(2)))/( x^(2)+1)` dx, is

Options

A. `I^(2)`
B. `(1)/(2) I`
C. `2I`
D. `(1)/(2) I^(2)`

Answer

To solve the integral \( J = \int_{0}^{1} \frac{x e^{x^2}}{x^2 + 1} \, dx \), we can use a substitution method. Let's go through the solution step-by-step.

### Step 1: Substitution
Let \( t = x^2 \). Then, differentiating both sides gives us:
\[ dt = 2x \, dx \quad \Rightarrow \quad \frac{dt}{2} = x \, dx \]

### Step 2: Change of Limits
When \( x = 0 \), \( t = 0^2 = 0 \).
When \( x = 1 \), \( t = 1^2 = 1 \).
Thus, the limits of integration remain the same: from \( 0 \) to \( 1 \).

### Step 3: Rewrite the Integral
Now, substituting \( t \) into the integral:
\[ J = \int_{0}^{1} \frac{x e^{x^2}}{x^2 + 1} \, dx = \int_{0}^{1} \frac{e^{t}}{t + 1} \cdot \frac{dt}{2} \]
This simplifies to:
\[ J = \frac{1}{2} \int_{0}^{1} \frac{e^{t}}{t + 1} \, dt \]

### Step 4: Recognize the Integral
Notice that the integral \( \int_{0}^{1} \frac{e^{t}}{t + 1} \, dt \) is the same as the integral \( I \) given in the problem:
\[ I = \int_{0}^{1} \frac{e^{x}}{x + 1} \, dx \]
Thus, we can write:
\[ J = \frac{1}{2} I \]

### Final Answer
Therefore, the value of the integral \( J \) is:
\[ J = \frac{I}{2} \]