Problem 154
Evaluate the integral: I = integral e^(x+ ln x) dx
Evaluate the integral: $$ I = \int e^{x+\ln x}\,dx. $$
Explanation of Method
Splitting the exponent as a sum lets $e^{\ln x}=x$ simplify the integrand to $xe^x$, which is then handled by a single integration by parts.
- Simplify the exponent using $e^{a+b}=e^ae^b$ and $e^{\ln x}=x$.
- Apply integration by parts to $\int xe^x\,dx$.
Step 1 · Simplify the integrand
Using $e^{x+\ln x}=e^x\cdot e^{\ln x}=e^x\cdot x$:
$$ I = \int xe^x\,dx. $$
Step 2 · Integration by parts
Let:
$$ u = x \quad \Rightarrow \quad du = dx, \qquad dv = e^x\,dx \quad \Rightarrow \quad v = e^x. $$
Applying the formula:
$$ I = xe^x - \int e^x\,dx. $$
Step 3 · Integrate
Since $\int e^x\,dx=e^x$:
$$ I = xe^x - e^x + C = e^x(x-1) + C. $$
Summary Table
| Quantity | Value |
|---|---|
| Simplified integrand | $xe^x$ |
| $u$ | $x$ |
| $dv$ | $e^x\,dx$ |
Additional Notes
- The domain requires $x>0$, where $\ln x$ is defined.
- Always check whether an exponent can be split and simplified before diving into integration by parts — here $e^{x+\ln x}$ looks intimidating but collapses to the elementary product $xe^x$ immediately.
- The result can be checked by differentiating $e^x(x-1)$ and confirming it returns $xe^x=e^{x+\ln x}$.
Result
The solution to the integral is:
$$
I = e^x(x-1) + C.
$$