`int(e^((x^(2)+4Inx))-x^(3)e^(x^(2)))/(x-1)dx` equals to

Options

A. `((e^(3 Inx)-e^(Inx))/(2x))e^(x^(2))+C`
B. `((x-1)xe^(x^(2)))/(2)+C`
C. `((x^(2)-1))/(2x)e^(x^(2))+C`
D. none of these

Explanation

To solve the integral

\[ I = \int \frac{e^{x^2 + 4 \ln x} - x^3 e^{x^2}}{x - 1} \, dx, \]

we can simplify the expression step by step.

### Step 1: Simplify the exponent
We know that \(4 \ln x = \ln(x^4)\). Therefore, we can rewrite the integral as:

\[ I = \int \frac{e^{x^2} \cdot e^{4 \ln x} - x^3 e^{x^2}}{x - 1} \, dx = \int \frac{e^{x^2} \cdot x^4 - x^3 e^{x^2}}{x - 1} \, dx. \]

### Step 2: Factor out common terms
Notice that both terms in the numerator contain \(e^{x^2}\):

\[ I = \int \frac{e^{x^2} (x^4 - x^3)}{x - 1} \, dx. \]

### Step 3: Factor the polynomial
We can factor \(x^4 - x^3\):

\[ x^4 - x^3 = x^3(x - 1). \]

Substituting this back into the integral gives:

\[ I = \int \frac{e^{x^2} \cdot x^3 (x - 1)}{x - 1} \, dx. \]

### Step 4: Cancel the common factor
The \(x - 1\) terms cancel out (for \(x \neq 1\)):

\[ I = \int x^3 e^{x^2} \, dx. \]

### Step 5: Substitution
Let \(u = x^2\). Then, \(du = 2x \, dx\) or \(dx = \frac{du}{2x}\). Since \(x = \sqrt{u}\), we have:

\[ dx = \frac{du}{2\sqrt{u}}. \]

Thus, we can rewrite \(x^3\) as \((\sqrt{u})^3 = u^{3/2}\):

\[ I = \int u^{3/2} e^u \cdot \frac{du}{2\sqrt{u}} = \frac{1}{2} \int u e^u \, du. \]

### Step 6: Integration by parts
Using integration by parts, let \(v = e^u\) and \(w = u\):

\[ \int u e^u \, du = u e^u - \int e^u \, du = u e^u - e^u + C. \]

### Step 7: Substitute back
Thus,

\[ I = \frac{1}{2} \left( u e^u - e^u \right) + C = \frac{1}{2} \left( x^2 e^{x^2} - e^{x^2} \right) + C. \]

### Step 8: Factor out \(e^{x^2}\)
Factoring out \(e^{x^2}\):

\[ I = \frac{e^{x^2}}{2} (x^2 - 1) + C. \]

### Final Answer
Thus, the final result of the integral is:

\[ I = \frac{e^{x^2}}{2} (x^2 - 1) + C. \]