Evaluate the integral (for n ≠-1): I = integral x^nln x dx

Evaluate the integral (for $n\neq-1$): $$ I = \int x^n\ln x\,dx. $$

Step 1 · Set up integration by parts

Let:

$$ u = \ln x \quad \Rightarrow \quad du = \frac{dx}{x}, \qquad dv = x^n\,dx \quad \Rightarrow \quad v = \frac{x^{n+1}}{n+1}, \quad n\neq-1. $$

Step 2 · Apply the formula

$$ I = \frac{x^{n+1}}{n+1}\ln x - \int \frac{x^{n+1}}{(n+1)x}\,dx = \frac{x^{n+1}}{n+1}\ln x - \frac{1}{n+1}\int x^n\,dx. $$

Step 3 · Integrate

Since $\int x^n\,dx=\dfrac{x^{n+1}}{n+1}$ for $n\neq-1$:

$$ I = \frac{x^{n+1}}{n+1}\ln x - \frac{x^{n+1}}{(n+1)^2} + C. $$

Summary Table

Additional Notes

• The domain requires $x>0$; the restriction $n\neq-1$ is needed because $\int x^{-1}dx=\ln|x|$ instead of a power of $x$, which changes the whole structure of the result (that case is Problem 106's family instead).
• Setting $n=1$ recovers Problem 143's result exactly, and setting $n=0$ recovers Problem 142's result, confirming the formula's consistency.
• The result can be checked by differentiating $\tfrac{x^{n+1}}{n+1}\ln x-\tfrac{x^{n+1}}{(n+1)^2}$ and confirming it returns $x^n\ln x$.