Question

Evaluate the following : `int (logx)2.dx`

Answer 1

Let I = `int (logx)^2.dx`
Put log x = t
∴ x = e^t
∴ dx = e^t dt
∴ I = `int t^2e^t dt`

= `t^2 int e^t dt - int [d/dx(t^2) int e^t - dt]dt`

= `t^2e^t - int 2te^t dt`

= `t^2e^t - 2[t int e^t dt - int {d/dt (t) int e^t dt}dt]`

= `t^2e^t - 2[te^t - int 1.e^t dt]`
= `t^2e^t - 2te^t + 2e^t + c`
= `e^t[t^2 - 2t + 2] + c`
= x[(log x)² – 2(log x) + 2] + c.
Alternative Method :
Let I = `int (logx)^2.dx`

= `int (logx)^2. 1dx`

= `(logx)^2 int1.dx - int[d/dx (logx)^2.int1.dx].dx`

= `(logx)^2.x - int 2logx.d/dx(logx).xdx`

= `x(logx)^2 - int 2logx xx 1/x xx x.dx`

= `x(logx)^2 - 2 int (logx).1dx`

= `x(logx)2 - 2[(logx) int 1.dx - int {d/dx (logx) int 1.dx}.dx]`

= `x(logx)^2 - 2[(logx)x - int1/x xx x.dx`

= `x(logx) - 2x(logx) + 2 int 1.dx`

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

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