If Im-n-xmln xndx - then m-1Im-n-nIm-n-1-where m-n-C is integration constant

Nbsp; I_m, n=∫ nbsp; x^m(lnx)^nd x =∫I 1(ln x)^n · nbsp; IIx^m Applying integration by parts I_m, n=1(lnx)^n· nbsp; x^m+1(m+1) ∫( n)(lnx)^n+1· nbsp; 1 ...

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Byju's Answer

Standard XII

Mathematics

Sum of n Terms

If Im,n=∫xmln...

Question

If $I_{m, n} = \int \frac{x^{m}}{(ln x)^{n}} d x,$ then $(m + 1) I_{m, n} - n I_{m, n + 1} =$
(where $m, n \in W; C$ is integration constant)

\frac{x^{m}}{(ln x)^{n}} + C

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- \frac{x^{m}}{(ln x)^{n}} + C

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- \frac{x^{m + 1}}{(ln x)^{n}} + C

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\frac{x^{m + 1}}{(ln x)^{n}} + C

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Solution

The correct option is D $\frac{x^{m + 1}}{(ln x)^{n}} + C$
$I_{m, n} = \int \frac{x^{m}}{(ln x)^{n}} d x = \int \frac{1}{(ln x)^{n}} I \cdot {x I I}^{m}$
Applying integration by parts
$I_{m, n} = \frac{1}{(ln x)^{n}} \cdot \frac{x^{m + 1}}{(m + 1)} - \int \frac{(- n)}{(ln x)^{n + 1}} \cdot \frac{1}{x} \cdot \frac{x^{m + 1}}{m + 1} d x = \frac{1}{(ln x)^{n}} \cdot \frac{x^{m + 1}}{(m + 1)} + \frac{n}{(m + 1)} \int \frac{x^{m}}{(ln x)^{n + 1}} d x \Rightarrow (m + 1) I_{m, n} = \frac{x^{m + 1}}{(ln x)^{n}} + {n I}_{m, n + 1} \Rightarrow (m + 1) I_{m, n} - n I_{m, n + 1} = \frac{x^{m + 1}}{(ln x)^{n}} + C$

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Q. lf

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Sum of n Terms

Standard XII Mathematics

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If Im,n=∫xm(lnx)ndx, then (m+1)Im,n−nIm,n+1= (where m,n∈W;C is integration constant)

The correct option is D xm+1(lnx)n+C Im,n=∫xm(lnx)ndx=∫1(lnx)nI⋅xIIm Applying integration by parts Im,n=1(lnx)n⋅xm+1(m+1)−∫(−n)(lnx)n+1⋅1x⋅xm+1m+1dx=1(lnx)n⋅xm+1(m+1)+n(m+1)∫xm(lnx)n+1dx⇒(m+1)Im,n=xm+1(lnx)n+nIm,n+1⇒(m+1)Im,n−nIm,n+1=xm+1(lnx)n+C

If $I_{m, n} = \int \frac{x^{m}}{(ln x)^{n}} d x,$ then $(m + 1) I_{m, n} - n I_{m, n + 1} =$
(where $m, n \in W; C$ is integration constant)