`intx2^(ln(x^(2)+1))dx` is equal to

Options

A. `(2^(ln(x^(2)+1)))/(2(x^(2)+1))+C`
B. `(x^(2)+1)2^(ln(x^(2)+1))/(ln2+1)`
C. `((x^(2)+1)^(ln2+1))/(2(ln2+1))+C`
D. `((x^(2)+1)^(ln2))/(2(ln2+1))+C`

Explanation

To solve the integral \( I = \int x \cdot 2^{\ln(x^2 + 1)} \, dx \), we can follow these steps:

### Step 1: Simplify the integrand
We start by rewriting \( 2^{\ln(x^2 + 1)} \) using the property of logarithms:
\[ 2^{\ln(x^2 + 1)} = (x^2 + 1)^{\ln 2} \]
Thus, the integral becomes:
\[ I = \int x \cdot (x^2 + 1)^{\ln 2} \, dx \]

### Step 2: Use substitution
Let \( t = x^2 + 1 \). Then, we differentiate to find \( dt \):
\[ dt = 2x \, dx \quad \Rightarrow \quad dx = \frac{dt}{2x} \]
Now, substituting \( x^2 = t - 1 \) gives us \( x = \sqrt{t - 1} \). Therefore, we can rewrite \( dx \) as:
\[ dx = \frac{dt}{2\sqrt{t - 1}} \]

### Step 3: Substitute into the integral
Substituting \( x \) and \( dx \) into the integral, we have:
\[ I = \int \sqrt{t - 1} \cdot t^{\ln 2} \cdot \frac{dt}{2\sqrt{t - 1}} = \frac{1}{2} \int t^{\ln 2} \, dt \]

### Step 4: Integrate
Now we can integrate:
\[ \frac{1}{2} \int t^{\ln 2} \, dt = \frac{1}{2} \cdot \frac{t^{\ln 2 + 1}}{\ln 2 + 1} + C \]

### Step 5: Substitute back for \( t \)
Now, substituting back \( t = x^2 + 1 \):
\[ I = \frac{1}{2(\ln 2 + 1)} (x^2 + 1)^{\ln 2 + 1} + C \]

### Final Answer
Thus, the final answer is:
\[ I = \frac{1}{2(\ln 2 + 1)} (x^2 + 1)^{\ln 2 + 1} + C \]

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