Question

Evaluate:

`int (logx)^2 dx`

Answer 1

I = `int (logx)^2*1dx`

Integrating by parts

I = `(logx)^2 int 1dx - int(int1dx d/dx(logx)^2)dx`

∴ I = `(logx)^2(x) - int(x*2logx*1/x)dx`

∴ I = `x(logx)^2 - 2intlogx*1dx`

Again integrating by parts,

I = `x(logx)^2 - 2[logx(x) - intx*1/xdx]`

= `x(logx)^2 - 2[xlogx - x] + c`

= `x(logx)^2 - 2xlogx + 2x + c`