Evaluate the integral: I = integral sqrt(x) ln²x dx
Evaluate the integral: $$ I = \int \sqrt{x}\,\ln^2x\,dx. $$
Explanation of Method
Integration by parts with $u=\ln^2x$, $dv=\sqrt x\,dx$ reduces the power of the logarithm by one, leaving a $\sqrt x\ln x$ integral that itself requires one more application of integration by parts.
- Choose $u=\ln^2x$, $dv=\sqrt x\,dx$ and apply integration by parts.
- Solve the leftover $\int \sqrt x\ln x\,dx$ with a second integration by parts.
- Substitute back and combine.
Step 1 · First integration by parts
Let:
$$ u = \ln^2x \quad \Rightarrow \quad du = \frac{2\ln x}{x}\,dx, \qquad dv = \sqrt x\,dx \quad \Rightarrow \quad v = \frac23x^{3/2}. $$
Applying the formula:
$$ I = \frac23x^{3/2}\ln^2x - \frac43\int x^{1/2}\ln x\,dx. $$
Step 2 · Second integration by parts
For $J=\displaystyle\int x^{1/2}\ln x\,dx$, let:
$$ u = \ln x \quad \Rightarrow \quad du = \frac{dx}{x}, \qquad dv = x^{1/2}\,dx \quad \Rightarrow \quad v = \frac23x^{3/2}. $$
Applying the formula:
$$ J = \frac23x^{3/2}\ln x - \frac23\int x^{1/2}\,dx = \frac23x^{3/2}\ln x - \frac49x^{3/2}. $$
Step 3 · Combine
Substituting $J$ back into the expression for $I$:
$$ I = \frac23x^{3/2}\ln^2x - \frac43\left(\frac23x^{3/2}\ln x - \frac49x^{3/2}\right). $$
Step 4 · Simplify
$$ I = \frac23x^{3/2}\ln^2x - \frac89x^{3/2}\ln x + \frac{16}{27}x^{3/2} + C. $$
Summary Table
| Piece | Value |
|---|---|
| First IBP | $\tfrac23x^{3/2}\ln^2x - \tfrac43J$ |
| $J=\int\sqrt x\ln x\,dx$ | $\tfrac23x^{3/2}\ln x - \tfrac49x^{3/2}$ |
Additional Notes
- The domain requires $x>0$.
- This "reduce the power of $\ln x$ by one at each application of integration by parts" strategy is the same pattern used in Problem 144, just with $v=\tfrac23x^{3/2}$ in place of $v=x^2/2$.
- The result can be checked by differentiating $\tfrac23x^{3/2}\ln^2x-\tfrac89x^{3/2}\ln x+\tfrac{16}{27}x^{3/2}$ and confirming it returns $\sqrt x\ln^2x$.
Result
The solution to the integral is: