Question

\(\int (x+1)e^x\,dx\) is equal to
[ \(C\) is an arbitrary constant]

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

Hint:

Use \(\int e^x(f+f')dx=e^xf\) with \(f=x\), or integrate by parts.

Answer 1

The Correct Option is C

Step 1: Choose the method.
The integrand is a product of a polynomial \(x+1\) and \(e^x\). Integration by parts is the natural tool. The formula is \(\int u\,dv=uv-\int v\,du\).

Step 2: Set u and dv.
Let \(u=x+1\) so \(du=dx\). Let \(dv=e^x dx\) so \(v=e^x\).

Step 3: Apply the formula.
\[ \int (x+1)e^x dx=(x+1)e^x-\int e^x dx=(x+1)e^x-e^x+C \]

Step 4: Simplify.
\[ (x+1)e^x-e^x=xe^x+e^x-e^x=xe^x \] So the integral is \(xe^x+C\).

Step 5: Check the options.
Option 1 is what we get before subtracting \(\int e^x dx\), so it forgets a term. Differentiating option 2 gives \(e^x\), not \((x+1)e^x\). Option 4 has no exponential and cannot work. Differentiating option 3 gives \(e^x+xe^x=(x+1)e^x\), which is correct.

Final Answer:
The integral is \(xe^x+C\), option 3. \[ \boxed{xe^x+C} \]