Question

lf $I_{m,n}=\int \frac{x^m}{(\log x{)}^n}dx$ then
$(m+1)I_{m,n}-n.I_{m,n+1}=$

• A. $\frac{x^m}{(\log x{)}^n}+c$
• B. $-\frac{x^m}{(\log x{)}^n}+c$
• C. $-\frac{x^{m+1}}{(\log x{)}^n}+c$
• D. $\frac{x^{m+1}}{(\log x{)}^n}+c$

Answer 1

The correct option is D $\frac{x^{m+1}}{(\log x{)}^n}+c$

$I_{m,n}=\int \frac{x^m}{(logx{)}^m}dx$

$I_{m,n}=\frac{1}{(logx{)}^n}\frac{x^{m+1}}{(m+1)}\int \frac{(-n)}{(logx{)}^{n+1}}\frac{1}{x}\frac{x^{n+1}}{m+1}dx$

$=\frac{1}{(logx{)}^n}-\frac{x^{m+1}}{(m+1)}+\frac{n}{(m+1)}\int \frac{x^m}{(logx{)}^{n+1}}dx$

$(m+1)I_{m,n}=[\frac{x^{m+1}}{(logx{)}^n}+n{I}_{m,n+1}]$

$(m+1)I_{m,n-n}{I}_{m,n+1}=\frac{x^{m+1}}{(logx{)}^n}+c$

$(m+1)I_{m,n-n}{I}_{m,n+1}=\frac{x^{m+1}}{(logx{)}^n}+c$