Question

14. x (logx)2

Answer 1

The integral is given as,

I= ∫ x ( logx ) 2 dx

Use integration by parts. Consider ( logx ) 2 as first function and xas second function

I= ∫ x ( logx ) 2 dx = ( logx ) 2 ∫ xdx − ∫ ( d dx ( logx ) 2 ∫ xdx ) dx = ( xlogx ) 2 2 − ∫ 2logx x × x 2 2 dx I= ( xlogx ) 2 2 − ∫ xlogxdx

Again apply integration by parts. Consider logarithmic function as first function and xas second function.

I= ( xlogx ) 2 2 −[ logx ∫ xdx − ∫ ( d dx logx ∫ xdx )dx ] = ( xlogx ) 2 2 −[ x 2 logx 2 − ∫ 1 x × x 2 2 dx ] I= ( xlogx ) 2 2 − x 2 logx 2 + 1 4 x 2 +C

Thus, the integration of ∫ x ( logx ) 2 dx is ( xlogx ) 2 2 − x 2 logx 2 + 1 4 x 2 +C.