Question

If \[ \int e^x\frac{x^3+3x^2+4}{(x+1)^3}\,dx = e^x f(x)+C, \] then \(f(x)\) is

• A. \[ \frac{x^2+2x-2}{(x+1)^2} \]
• B. \[ \frac{x^2+x-1}{(x+1)^2} \]
• C. \[ \frac{x^2-2x+2}{(x+1)^2} \]
• D. \[ \frac{x^2+2x-1}{(x+1)^2} \] \bigskip

Hint:

For integrals of the form \[ \int e^x g(x)\,dx=e^x f(x)+C, \] always use \[ f+f'=g(x). \] This avoids repeated integration by parts.

Answer 1

The Correct Option is A

Concept: If \[ \int e^x g(x)\,dx=e^x f(x)+C, \] then differentiating gives \[ e^x(f+f')=e^x g(x), \] hence \[ f+f'=g(x). \]

Step 1: Identify \(g(x)\). \[ g(x)=\frac{x^3+3x^2+4}{(x+1)^3}. \] We test the options using \[ f+f'=g(x). \]

Step 2: Take Option (A). \[ f(x)=\frac{x^2+2x-2}{(x+1)^2}. \] Differentiate: \[ f'(x) = \frac{(2x+2)(x+1)^2-2(x+1)(x^2+2x-2)} {(x+1)^4}. \] Factor \((x+1)\): \[ f'(x) = \frac{(x+1)\Big[(2x+2)(x+1)-2(x^2+2x-2)\Big]} {(x+1)^4}. \] \[ = \frac{2x^2+4x+2-2x^2-4x+4} {(x+1)^3}. \] \[ = \frac{6}{(x+1)^3}. \] Therefore, \[ f+f' = \frac{(x^2+2x-2)(x+1)+6} {(x+1)^3}. \] Expanding, \[ (x^2+2x-2)(x+1) = x^3+3x^2-2. \] Hence \[ f+f' = \frac{x^3+3x^2-2+6} {(x+1)^3}. \] \[ = \frac{x^3+3x^2+4} {(x+1)^3}. \] \[ =f(x). \] Thus Option (A) satisfies the required condition.

Step 3: Write the final answer. \[ \boxed{ f(x)= \frac{x^2+2x-2}{(x+1)^2} } \]