Evaluate the integral: I = integral x³ln²x dx
Evaluate the integral: $$ I = \int x^3\ln^2x\,dx. $$
Explanation of Method
Integration by parts with $u=\ln^2x$, $dv=x^3\,dx$ reduces the power of the logarithm by one, leaving a $\int x^3\ln x\,dx$ integral that itself requires a second application of integration by parts.
- Choose $u=\ln^2x$, $dv=x^3\,dx$ and apply integration by parts.
- Solve the leftover $\int x^3\ln x\,dx$ with a second integration by parts.
- Combine the results.
Step 1 · First integration by parts
Let:
$$ u = \ln^2x \quad \Rightarrow \quad du = \frac{2\ln x}{x}\,dx, \qquad dv = x^3\,dx \quad \Rightarrow \quad v = \frac{x^4}{4}. $$
Applying the formula:
$$ I = \frac{x^4}{4}\ln^2x - \frac12\int x^3\ln x\,dx. $$
Step 2 · Second integration by parts
For $\displaystyle\int x^3\ln x\,dx$, let $u=\ln x$, $dv=x^3\,dx$, so $du=dx/x$, $v=x^4/4$:
$$ \int x^3\ln x\,dx = \frac{x^4}{4}\ln x - \frac14\int x^3\,dx = \frac{x^4}{4}\ln x - \frac{x^4}{16}. $$
Step 3 · Combine
Substituting back:
$$ I = \frac{x^4}{4}\ln^2x - \frac12\left(\frac{x^4}{4}\ln x - \frac{x^4}{16}\right). $$
Step 4 · Simplify
$$ I = \frac{x^4}{4}\ln^2x - \frac{x^4}{8}\ln x + \frac{x^4}{32} + C. $$
Summary Table
| Piece | Value |
|---|---|
| First IBP | $\tfrac{x^4}{4}\ln^2x - \tfrac12\int x^3\ln x\,dx$ |
| $\int x^3\ln x\,dx$ | $\tfrac{x^4}{4}\ln x - \tfrac{x^4}{16}$ |
Additional Notes
- The domain requires $x>0$.
- This follows the same reduce-the-power-of-$\ln x$ pattern as Problems 144 and 149, using $v=x^4/4$ in place of $x^2/2$ or $\tfrac23x^{3/2}$.
- The result can be checked by differentiating $\tfrac{x^4}{4}\ln^2x-\tfrac{x^4}{8}\ln x+\tfrac{x^4}{32}$ and confirming it returns $x^3\ln^2x$.
Result
The solution to the integral is: