Question

The value of the definite integral \[ \int_{0}^{1}\frac{xe^{x}}{(x+1)^2}\,dx \] is equal to

• A. \(\dfrac{e-2}{2}\)
• B. \(\dfrac{e}{2}\)
• C. \(\dfrac{e-1}{2}\)
• D. \(\dfrac{e-3}{2}\)

Hint:

Whenever the integrand contains \[ \frac{xe^x}{(x+1)^2}, \] check whether it is \[ \boxed{ \frac{d}{dx} \left( \frac{e^x}{x+1} \right) } \] This makes the integration immediate.

Answer 1

The Correct Option is A

Concept:

Notice that the integrand is the derivative of a simple quotient.

Step 1: Differentiate the function

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

Using the quotient rule,

\[ \begin{aligned} \frac{d}{dx}\left(\frac{e^x}{x+1}\right) &= \frac{e^x(x+1)-e^x}{(x+1)^2} \\ &= \frac{xe^x}{(x+1)^2}. \end{aligned} \]

Hence,

\[ \frac{xe^x}{(x+1)^2} = \frac{d}{dx}\left(\frac{e^x}{x+1}\right). \]

Step 2: Apply the Fundamental Theorem of Calculus.

\[ \begin{aligned} I &= \int_0^1 \frac{xe^x}{(x+1)^2}\,dx \\ &= \left[\frac{e^x}{x+1}\right]_0^1 \\ &= \frac{e}{2}-1 \\ &= \frac{e-2}{2}. \end{aligned} \]

Therefore,

\[ \int_0^1 \frac{xe^x}{(x+1)^2}\,dx = \frac{e-2}{2}. \]

Hence, the correct option is (A).