Problem 152
Evaluate the integral: I = integral (ln x)/(x³) dx
Evaluate the integral: $$ I = \int \frac{\ln x}{x^3}\,dx. $$
Explanation of Method
Integration by parts with $u=\ln x$, $dv=x^{-3}\,dx$ leaves a leftover power-rule integral that resolves directly.
- Choose $u=\ln x$, $dv=x^{-3}\,dx$.
- Differentiate $u$ and integrate $dv$.
- Apply the integration by parts formula and simplify.
Step 1 · Set up integration by parts
Let:
$$ u = \ln x \quad \Rightarrow \quad du = \frac{dx}{x}, \qquad dv = x^{-3}\,dx \quad \Rightarrow \quad v = -\frac{1}{2x^2}. $$
Step 2 · Apply the formula
$$ I = -\frac{\ln x}{2x^2} - \int \left(-\frac{1}{2x^2}\right)\frac{dx}{x} = -\frac{\ln x}{2x^2} + \frac12\int x^{-3}\,dx. $$
Step 3 · Integrate the leftover term
Since $\int x^{-3}\,dx = -\dfrac{1}{2x^2}$:
$$ I = -\frac{\ln x}{2x^2} + \frac12\left(-\frac{1}{2x^2}\right) + C. $$
Step 4 · Simplify
$$ I = -\frac{\ln x}{2x^2} - \frac{1}{4x^2} + C. $$
Summary Table
| Quantity | Value |
|---|---|
| $u$ | $\ln x$ |
| $dv$ | $x^{-3}\,dx$ |
| Antiderivative | $-\tfrac{\ln x}{2x^2}-\tfrac{1}{4x^2}$ |
Additional Notes
- The domain requires $x>0$.
- This is the negative-exponent counterpart of Problem 145's general formula $\int x^n\ln x\,dx$ for $n\neq-1$, here with $n=-3$.
- The result can be checked by differentiating $-\tfrac{\ln x}{2x^2}-\tfrac{1}{4x^2}$ and confirming it returns $\ln x/x^3$.
Result
The solution to the integral is:
$$
I = -\frac{\ln x}{2x^2} - \frac{1}{4x^2} + C.
$$