Question

Evaluate the integral: $\int e^{(e^x + x)} dx$

• A. $e^x + x + c$
• B. $e^{(e^x)} \cdot x + c$
• C. $e^{(e^x)} + c$
• D. $e^{(e^x)}(e^x - 1) + c$

Hint:

Whenever you see a compound exponential function like $e^{(e^x)}$, applying the chain rule to take its derivative immediately gives $\frac{d}{dx}\left(e^{(e^x)}\right) = e^{(e^x)} \cdot e^x = e^{(e^x + x)}$. Because differentiation and integration are inverse operations, noticing this derivative loop instantly proves that option (C) is the right answer without writing out full substitution steps.

Answer 1

The Correct Option is C

Step 1: Break apart the exponent.
Use $e^{A+B} = e^A \cdot e^B$ to write $e^{e^x + x} = e^{e^x}\cdot e^x$. So the integral becomes $\displaystyle\int e^{e^x}\cdot e^x\,dx$.
Step 2: Spot the inner-outer pattern.
Notice $e^x$ is exactly the derivative of $e^x$, the quantity sitting in the upper exponent. This is the classic $\int e^{f(x)}f'(x)\,dx$ shape.
Step 3: Substitute.
Let $t = e^x$. Then $dt = e^x\,dx$, which is precisely the leftover factor in the integrand.
Step 4: Rewrite the integral in $t$.
The integral turns into $\displaystyle\int e^t\,dt$.
Step 5: Integrate.
$\displaystyle\int e^t\,dt = e^t + c$.
Step 6: Restore the original variable.
Replacing $t = e^x$ gives $e^{e^x} + c$.
\[ \boxed{e^{e^x} + c} \]