Question

Given I_m =$\int_{1}^{e}{(\log x)^{m}}$dx. If $\frac{I_{m}}{K} + \frac{I_{m - 2}}{L}$ = e then values of K and L are –

• A. 1 – m, $\frac{1}{m}$
• B. $\frac{1}{1 - m}$, m
• C. $\frac{1}{1 - m}$, $\frac{m(m - 2)}{m - 1}$
• D. $\frac{m}{m - 1}$, m – 2

Answer 1

Answer: (A)

1 – m, $\frac{1}{m}$

Answer 2

I_m = $\int_{1}^{e}{(\log x)^{m}}$dx = x$\left. \ (\log x)^{m} \right|_{1}^{e}$

– m$\int_{1}^{e}{(\log x)^{m - 1}dx}$

= e – m $\left\lbrack \left. \ x(\log x)^{m - 1} \right|_{1}^{e} - (m - 1)\int_{1}^{e}{(\log x)^{m - 2}dx} \right\rbrack$

= e – me + m (m – 1) I_m–2 = (1 – m) e + m (m – 1) I_m–2

So $\frac{I_{m}}{1 - m}$ + m I_{m– 2} = e. Thus K = 1 – m and L = $\frac{1}{m}$.