Question

I_m,n = $\int_{0}^{1}{x^{m}(\mathcal{l}nx)^{n}dx}$

• A. $–\frac{n}{m}$I_m,n–1
• B. – $\frac{n}{m + 1}$I_m,n–1
• C. – $\frac{n + 1}{m}$I_m–1,n–1
• D. None of these

Answer 1

Answer: (B)

– $\frac{n}{m + 1}$I_m,n–1

Answer 2

I_m,n = $\int_{0}^{1}{x^{m}(\log x)^{n}dx}$

=$\left\lbrack (\log x)^{n}.\frac{x^{m + 1}}{m + 1} \right\rbrack_{0}^{1}$–$\int_{0}^{1}{n(\log x)^{n–1}}$.

$\frac{1}{x}$. $\frac{x^{m + 1}}{m + 1}$dx

= 0 – $\frac{n}{m + 1}$ $\int_{0}^{1}{x^{m}.(\log x)^{n–1}dx}$

= – $\frac{n}{m + 1}$ I_m,n–1