Evaluate the integral: I = integral (ln x)/x dx
Evaluate the integral: $$ I = \int \frac{\ln x}{x} \, dx. $$
Explanation of Method
Applying integration by parts with $u=\ln x$ and $dv=\frac{dx}{x}$ produces the original integral $I$ again on the right-hand side. Rather than being circular, this lets us solve for $I$ algebraically: the equation $I = \ln^2x - I$ rearranges to $2I=\ln^2x$.
- Apply integration by parts with $u=\ln x$, $dv=\frac{dx}{x}$.
- Recognize that the resulting integral is $I$ itself.
- Solve the resulting equation for $I$ algebraically.
- Add the constant of integration.
Step 1 · Set up integration by parts
To solve the integral, we use the technique of integration by parts. The formula for integration by parts is given by:
$$ \int u\,dv = uv - \int v\,du. $$
We choose:
$$ u = \ln x \quad \text{and} \quad dv = \frac{dx}{x}. $$
Now, compute the derivative and antiderivative:
$$ du = \frac{1}{x}\,dx \quad \text{and} \quad v = \ln x. $$
Step 2 · Apply the formula
Substituting into the formula for integration by parts:
$$ I = \int \frac{\ln x}{x}\,dx = \ln x\cdot \ln x - \int \ln x\cdot \frac{1}{x}\,dx = \ln^2 x - \int \frac{\ln x}{x}\,dx. $$
Notice that the remaining integral on the right is exactly $I$ again:
$$ I = \ln^2 x - I. $$
Step 3 · Solve for $I$
Adding $I$ to both sides:
$$ 2I = \ln^2 x. $$
Dividing both sides by $2$:
$$ I = \frac{\ln^2 x}{2} + C. $$
Summary Table
| Quantity | Value |
|---|---|
| $u$, $dv$ | $\ln x,\ dx/x$ |
| $du$, $v$ | $dx/x,\ \ln x$ |
| Self-referential equation | $I = \ln^2x - I$ |
Additional Notes
- The domain of $I$ requires $x > 0$, since $\ln x$ is only defined there.
- A quicker route to the same result is the direct substitution $u=\ln x$, $du=dx/x$, giving $I=\int u\,du = u^2/2+C$ — the self-referential integration-by-parts trick shown here is a useful technique in its own right, especially for higher powers of $\ln x$ (see Problem 85), but is not strictly necessary in this simplest case.
- The result can be checked by differentiating $\dfrac{\ln^2x}{2}$ with the chain rule and confirming it returns $\dfrac{\ln x}{x}$.
Result
The solution to the integral is: