`int(x e^x)/((1+x)^2)dx` is equal to

Options

A. `(e^(x))/(x+1)+C`
B. `e^(x)(x+1)+C`
C. `-(e^(x))/((x+1)^(2))+C`
D. `(e^(x))/(1+x^(2))+C`

Answer

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

### Step 1: Rewrite the Integral
We start by rewriting \( x \) in terms of \( (1+x) \):
\[ x = (1+x) - 1 \]
Thus, we can express the integral as:
\[ I = \int \frac{((1+x) - 1)e^x}{(1+x)^2} \, dx = \int \frac{(1+x)e^x}{(1+x)^2} \, dx - \int \frac{e^x}{(1+x)^2} \, dx \]
This simplifies to:
\[ I = \int \frac{e^x}{1+x} \, dx - \int \frac{e^x}{(1+x)^2} \, dx \]

### Step 2: Integration by Parts
Now we will use integration by parts on the first integral \( \int \frac{e^x}{1+x} \, dx \). Let:
- \( u = \frac{1}{1+x} \) → \( du = -\frac{1}{(1+x)^2} \, dx \)
- \( dv = e^x \, dx \) → \( v = e^x \)

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

### Step 3: Substitute Back
Now we substitute this back into our expression for \( I \):
\[ I = \left( \frac{e^x}{1+x} + \int \frac{e^x}{(1+x)^2} \, dx \right) - \int \frac{e^x}{(1+x)^2} \, dx \]
The two integrals \( \int \frac{e^x}{(1+x)^2} \, dx \) cancel out:
\[ I = \frac{e^x}{1+x} + C \]

### Final Answer
Thus, the solution to the integral is:
\[ I = \frac{e^x}{1+x} + C \]