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Question

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


A
xm(logx)n+c
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B
xm(logx)n+c
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C
xm+1(logx)n+c
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D
xm+1(logx)n+c
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Solution

The correct option is D $$\displaystyle \frac{x^{m+1}}{(\log x)^{n}}+c$$
$$\displaystyle I_{m,n}=\int \dfrac{x^{m}}{(log\ x)^{m}}dx$$
$$\displaystyle I_{m,n}=\dfrac{1}{(log\ x)^{n}}\dfrac{x^{m+1}}{(m+1)}\int \dfrac{(-n)}{(log\ x)^{n+1}}\dfrac{1}{x}\dfrac{x^{n+1}}{m+1}dx$$
$$\displaystyle =\dfrac{1}{(log\ x)^{n}}-\dfrac{x^{m+1}}{(m+1)}+\dfrac{n}{(m+1)}\int \dfrac{x^{m}}{(log x)^{n+1}}dx$$
$$\displaystyle (m+1)I_{m,n}=\left [ \dfrac{x^{m+1}}{(log\ x)^{n}}+n I_{m,n+1} \right ]$$
$$\displaystyle (m+1)I_{m,n-n} I_{m,n+1}=\dfrac{x^{m+1}}{(log\ x)^{n}}+c$$
$$\displaystyle (m+1)I_{m,n-n}I_{m,n+1}=\dfrac{x^{m+1}}{(log\ x)^{n}}+c$$

Mathematics

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