Question

Evaluate the integral \[ \int e^x (x+1)^2 dx. \]

• A. \( xe^x + c \)
• B. \( e^x x^2 + c \)
• C. \( e^x (x^2 + 1) + c \)
• D. \( e^x (x + 1) + c \)

Hint:

For integrals involving \( e^x \) multiplied by polynomials, use integration by parts recursively.

Answer 1

The Correct Option is C

Step 1: Expanding the function Expanding \( (x+1)^2 \): \[ (x+1)^2 = x^2 + 2x + 1. \] Rewriting the integral: \[ I = \int e^x (x^2 + 2x + 1) dx. \] Step 2: Splitting into separate integrals \[ I = \int e^x x^2 dx + 2\int e^x x dx + \int e^x dx. \] Using integration by parts, where \( u = x^2 \), \( dv = e^x dx \): \[ du = 2x dx, \quad v = e^x. \] Applying integration by parts repeatedly: \[ \int x^2 e^x dx = e^x (x^2 - 2x + 2). \] Similarly, \[ \int x e^x dx = e^x (x - 1). \] Step 3: Evaluating and summing terms \[ I = e^x (x^2 + 1) + c. \]

Answer 2

Evaluate the integral:
\[ I = \int e^x (x+1)^2 \, dx \]

Step 1: Expand the square:
\[ (x+1)^2 = x^2 + 2x + 1 \] So, \[ I = \int e^x (x^2 + 2x + 1) \, dx = \int e^x x^2 \, dx + 2 \int e^x x \, dx + \int e^x \, dx \]

Step 2: Evaluate each integral separately using integration by parts.

- For \( \int e^x x^2 \, dx \):
Let \( u = x^2 \), \( dv = e^x dx \) then \( du = 2x dx \), \( v = e^x \).
\[ \int e^x x^2 dx = x^2 e^x - \int 2x e^x dx \]
- For \( \int 2x e^x dx \):
Let \( u = 2x \), \( dv = e^x dx \), then \( du = 2 dx \), \( v = e^x \).
\[ \int 2x e^x dx = 2x e^x - \int 2 e^x dx = 2x e^x - 2 e^x + C \]

Step 3: Substitute back:
\[ \int e^x x^2 dx = x^2 e^x - (2x e^x - 2 e^x) = x^2 e^x - 2x e^x + 2 e^x \]

Step 4: Now evaluate the entire integral \( I \):
\[ I = \left(x^2 e^x - 2x e^x + 2 e^x\right) + \left(2x e^x - 2 e^x\right) + e^x + C \] Simplify:
\[ I = x^2 e^x - 2x e^x + 2 e^x + 2x e^x - 2 e^x + e^x + C = x^2 e^x + e^x + C = e^x (x^2 + 1) + C \]

Therefore, the value of the integral is:
\[ \boxed{e^x (x^2 + 1) + C} \]