Answer 1

`∫(log x)^2 dx`

let `u = (logx)^2 , "v" = 1`

`∫u."v" dx = u∫"v"dx - ∫[(du)/dx∫"v"dx]dx`

`therefore ∫ (log x)^2 . 1dx = (log x)^2 ∫1dx - ∫[2log x xx 1/x xx xdx]`

 = `x(log|x|^2) - 2∫log x  dx`

`x(log x)^2 - 2(x log|x| - x) + C`

 = `x(log|x|)^2 - 2x log|x| + 2x + C` .