`int(xe^x)/((x+1)^2)dx =`

Options

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

Answer

To solve the integral

\[ I = \int \frac{x e^x}{(x+1)^2} \, dx, \]

we will use substitution and integration by parts. Here’s a step-by-step solution:

### Step 1: Substitution
Let \( t = x + 1 \). Then, we have:
\[ x = t - 1 \quad \text{and} \quad dx = dt. \]

### Step 2: Rewrite the Integral
Substituting \( x \) and \( dx \) into the integral, we get:
\[ I = \int \frac{(t - 1)e^{t - 1}}{t^2} \, dt. \]

### Step 3: Simplify the Integral
We can separate the integral:
\[ I = \int \frac{(t - 1)e^{t - 1}}{t^2} \, dt = \int \frac{t e^{t - 1}}{t^2} \, dt - \int \frac{e^{t - 1}}{t^2} \, dt. \]
This simplifies to:
\[ I = \int \frac{e^{t - 1}}{t} \, dt - \int \frac{e^{t - 1}}{t^2} \, dt. \]

### Step 4: Integration by Parts
Now, we will solve the first integral using integration by parts. Let:
- \( u = \frac{1}{t} \) so that \( du = -\frac{1}{t^2} dt \),
- \( dv = e^{t - 1} dt \) so that \( v = e^{t - 1} \).

Using the integration by parts formula \( \int u \, dv = uv - \int v \, du \), we have:
\[ \int \frac{e^{t - 1}}{t} \, dt = \frac{e^{t - 1}}{t} - \int e^{t - 1} \left(-\frac{1}{t^2}\right) dt. \]
This gives us:
\[ \int \frac{e^{t - 1}}{t} \, dt = \frac{e^{t - 1}}{t} + \int \frac{e^{t - 1}}{t^2} \, dt. \]

### Step 5: Combine the Integrals
Now substituting back into the expression for \( I \):
\[ I = \left( \frac{e^{t - 1}}{t} + \int \frac{e^{t - 1}}{t^2} \, dt \right) - \int \frac{e^{t - 1}}{t^2} \, dt. \]
The integrals \( \int \frac{e^{t - 1}}{t^2} \, dt \) cancel out:
\[ I = \frac{e^{t - 1}}{t}. \]

### Step 6: Substitute Back
Now substituting back \( t = x + 1 \):
\[ I = \frac{e^{(x + 1) - 1}}{x + 1} = \frac{e^x}{x + 1}. \]

### Final Answer
Thus, the solution to the integral is:
\[ \int \frac{x e^x}{(x + 1)^2} \, dx = \frac{e^x}{x + 1} + C, \]
where \( C \) is the constant of integration.