Question

The value of \(\int\frac{xe^x dx}{(1+x)^2}\) is equal to

• A. e^x(1 + x) + c
• B. e^x(1 + x²) + c
• C. e^x(1 + x)² + c
• D. \(\frac{e^x}{1+x}+c\)

Answer 1

The Correct Option is D

We are given the integral \( \int \frac{e^x}{1+x} \, dx \). We will use integration by parts.

Let:

• \( u = \frac{1}{1+x} \), so \( du = -\frac{1}{(1+x)^2} dx \)
• \( dv = e^x dx \), so \( v = e^x \)

Using the formula for integration by parts, \( \int u \, dv = uv - \int v \, du \), we get:

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

Now, simplifying the expression:

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

At this point, we notice that the integral \( \int \frac{e^x}{(1+x)^2} \, dx \) appears again. Therefore, we can isolate the original integral on one side:

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

Now, substitute the value of the second integral back into the equation:

\[ \int \frac{e^x}{1+x} \, dx = \frac{e^x}{1+x} + C \]

Thus, the value of the integral is:

\[ \int \frac{e^x}{1+x} \, dx = \frac{e^x}{1+x} + C \]

Answer 2

We are given the integral:

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

Step 1: Use substitution

Let \( u = 1 + x \), so that \( du = dx \), and \( x = u - 1 \). The integral becomes:

\[ \int \frac{(u - 1) e^{u - 1}}{u^2} \, du \]

Step 2: Simplify the expression

Since \( e^{u - 1} = \frac{e^u}{e} \), the integral becomes:

\[ \frac{1}{e} \int \left( \frac{e^u}{u} - \frac{e^u}{u^2} \right) \, du \]

Step 3: Solve the integrals

The integral of \( \frac{e^u}{u^2} \) is \( -\frac{e^u}{u} \), and after simplification, the result is:

\[ \frac{e^x}{1 + x} + c \]

The correct answer is (D): \(\frac{e^x}{1+x} + c\).