Evaluate the integral: I = integral (ln³ x)/x dx
Evaluate the integral: $$ I = \int \frac{\ln^3 x}{x} \, dx. $$
Explanation of Method
As in Problem 84, integration by parts with $dv=\frac{dx}{x}$ produces a multiple of the original integral on the right-hand side. Here the coefficient is $3$ instead of $1$, so the self-referential equation becomes $I=\ln^4x-3I$, which still solves cleanly for $I$.
- Apply integration by parts with $u=\ln^3x$, $dv=\frac{dx}{x}$.
- Recognize that the resulting integral is a multiple of $I$.
- Solve the resulting equation for $I$ algebraically.
- Add the constant of integration.
Step 1 · Set up integration by parts
We start by choosing:
$$ u = \ln^3 x \quad \text{and} \quad dv = \frac{dx}{x}. $$
Now, compute the derivative and antiderivative:
$$ du = 3\ln^2x\cdot \frac{1}{x}\,dx \quad \text{and} \quad v = \ln x. $$
Step 2 · Apply the formula
Substituting into the integration by parts formula:
$$ I = \int \frac{\ln^3x}{x}\,dx = \ln^3x\cdot \ln x - \int \ln x\cdot 3\ln^2 x\cdot \frac{dx}{x}. $$
Simplifying:
$$ I = \ln^4 x - 3\int \frac{\ln^3 x}{x}\,dx = \ln^4x - 3I. $$
Step 3 · Solve for $I$
Collect the terms:
$$ 4I = \ln^4 x. $$
Finally, divide both sides by $4$ to solve for the integral:
$$ I = \frac{\ln^4 x}{4} + C. $$
Summary Table
| Quantity | Value |
|---|---|
| $u$, $dv$ | $\ln^3x,\ dx/x$ |
| Self-referential equation | $I = \ln^4x - 3I$ |
| Solved | $4I = \ln^4x$ |
Additional Notes
- The domain of $I$ requires $x>0$, since $\ln x$ is only defined there.
- This same technique generalizes: for $\displaystyle \int \frac{\ln^n x}{x}\,dx$, integration by parts always yields $I = \ln^{n+1}x - nI$, so $I = \dfrac{\ln^{n+1}x}{n+1}+C$ — matching the direct substitution $u=\ln x$ result, but arrived at via the self-referential route.
- The result can be checked by differentiating $\dfrac{\ln^4x}{4}$ with the chain rule and confirming it returns $\dfrac{\ln^3x}{x}$.
Result
The solution to the integral is: