If ${I_n} = \smallint \,{(\log \,\,x)^n}\;\,\,dx,\;\,{\rm{then}}\,{I_n}\,\, + \,\,n{I_{n - 1}}$ is equal to

${\left( {x\log \,\,x} \right)^n}$

${\left( {\log \,\,x} \right)^{n - 1}}$

$n\,\,{\left( {\log \,\,x} \right)^n}$

JEE MATHEMATICS MCQ: INDEFINITE INTEGRATION

1 Easy

1 of 1

Application +4 | -1

{I_n} = \int \,{(\log \,\,x)^n}\;\,\,dx,\;\,\mathrm{then}\,{I_n}\,\, + \,\,n{I_{n - 1}}

is equal to

View answer and explanation

Correct option: A

x\,{(\log \,\,x)^n}

Option A (

x{(\log \\,x)^n}

) is correct because integrating by parts simplifies the expression to

x{(\log \\,x)^n}