Evaluate the integral: integral (e^(1/x))/(x²) dx
Evaluate the integral: $$ \int \frac{e^{\frac{1}{x}}}{x^2} \, dx. $$
Explanation of Method
The derivative of $\tfrac1x$ is $-\tfrac{1}{x^2}$, which matches the $\tfrac{1}{x^2}$ factor in the integrand up to a sign. Substituting $u=\tfrac1x$ therefore turns the integral into $-\int e^u\,du$, a standard exponential antiderivative.
- Substitute $u=\tfrac1x$ to match the exponent.
- Convert $\tfrac{1}{x^2}\,dx$ into $-du$.
- Integrate the resulting exponential in $u$.
- Substitute back and add the constant of integration.
Step 1 · Substitution
We solve this integral using substitution. Let:
$$ u = \frac{1}{x}. $$
Then, differentiate both sides with respect to $x$:
$$ \frac{du}{dx} = -\frac{1}{x^2} \quad \Rightarrow \quad du = -\frac{1}{x^2}\,dx. $$
Step 2 · Substitute and integrate
Thus, we can substitute $du = -\dfrac{1}{x^2}\,dx$ into the integral:
$$ \int \frac{e^{1/x}}{x^2}\,dx = \int e^u\cdot(-du) = -\int e^u\,du. $$
The integral of $e^u$ is simply $e^u$, so we have:
$$ -e^u. $$
Step 3 · Substitute back
Finally, substitute back $u = \dfrac1x$ to return to the variable $x$:
$$ -e^{1/x} + C. $$
Summary Table
| Quantity | Value |
|---|---|
| Substitution | $u = 1/x$ |
| $dx/x^2$ in terms of $du$ | $-du$ |
| Antiderivative | $-e^u$ |
Additional Notes
- The domain of $I$ requires $x \neq 0$, since $1/x$ and $1/x^2$ are undefined there.
- The negative sign in $du=-\tfrac{1}{x^2}dx$ flips the sign of the resulting antiderivative — a small but easy detail to miss when the factor of $1/x^2$ is written with an implicit positive sign in the integrand.
- The result can be checked by differentiating $-e^{1/x}$ with the chain rule and confirming it returns $\dfrac{e^{1/x}}{x^2}$.
Result
The solution to the integral is: