Evaluate the integral: I = integral (ln²x)/(x²) dx
Evaluate the integral: $$ I = \int \frac{\ln^2x}{x^2}\,dx. $$
Explanation of Method
Substituting $x=e^u$ (so $u=\ln x$) turns the integrand into $u^2e^{-u}$, a standard polynomial-times-exponential integral that is resolved by two applications of integration by parts.
- Substitute $u=\ln x$, i.e. $x=e^u$.
- Apply integration by parts twice to $\int u^2e^{-u}\,du$.
- Substitute back and simplify.
Step 1 · Substitution
Let $u=\ln x$, so $x=e^u$ and $dx=e^u\,du$. Then:
$$ \frac{\ln^2x}{x^2}\,dx = \frac{u^2}{e^{2u}}\cdot e^u\,du = u^2e^{-u}\,du. $$
So the integral becomes:
$$ I = \int u^2e^{-u}\,du. $$
Step 2 · First integration by parts
Let $v=u^2$, $dw=e^{-u}du$, so $dv=2u\,du$, $w=-e^{-u}$:
$$ \int u^2e^{-u}\,du = -u^2e^{-u} + \int 2ue^{-u}\,du. $$
Step 3 · Second integration by parts
For $\displaystyle\int 2ue^{-u}\,du$, let $v=2u$, $dw=e^{-u}du$, so $dv=2\,du$, $w=-e^{-u}$:
$$ \int 2ue^{-u}\,du = -2ue^{-u} + \int 2e^{-u}\,du = -2ue^{-u} - 2e^{-u}. $$
Step 4 · Combine and substitute back
Combining:
$$ \int u^2e^{-u}\,du = -u^2e^{-u} - 2ue^{-u} - 2e^{-u} + C = -\left(u^2+2u+2\right)e^{-u} + C. $$
Substituting back $u=\ln x$, $e^{-u}=1/x$:
$$ I = -\frac{\ln^2x}{x} - \frac{2\ln x}{x} - \frac2x + C. $$
Summary Table
| Piece | Value |
|---|---|
| Substitution | $u=\ln x,\;x=e^u$ |
| Reduced integral | $\int u^2e^{-u}\,du$ |
| Antiderivative | $-(u^2+2u+2)e^{-u}$ |
Additional Notes
- The domain requires $x>0$.
- A common bookkeeping slip in the two-stage integration by parts here is losing a minus sign on the second application (since $\int e^{-u}du=-e^{-u}$, not $+e^{-u}$); it's worth double-checking each $w=-e^{-u}$ substitution carries through consistently.
- The result can be checked by differentiating $-\tfrac{\ln^2x}{x}-\tfrac{2\ln x}{x}-\tfrac2x$ and confirming it returns $\ln^2x/x^2$.
Result
The solution to the integral is: