Question

If x = Φ(t) is a differentiable function of t, then prove that:

`int f(x)dx = int f[Φ(t)]*Φ^'(t)dt`

Hence, find `int(logx)^n/x dx`.

Answer 1

x = Φ(t) is differentiable function of t.

∴ `dx/dt` = Φ'(t)

Let `intf(x)dx` = F(x)

∴ `d/dx[F(x)]` = f(x)

∴ By the chain rule

`d/dt[F(x)] = d /dx[F(x)]*dx/dt`

=  `f(x)*dx/dt`

= `f[Φ(t)]*Φ^'(t)`

∴ By the definition of integral F(x) = `int f[Φ(t)]*Φ^'(t)dt`

∴ `int f(x)dx = int f[Φ(t)]*Φ^'(t)dt`

To find `bb(int (log  x)^n/x dx)`:

Let I = `int (logx)^n/x dx`

Put log x = t

∴ `1/x dx` = dt

∴ I = `int t^n dt = t^(n + 1)/(n + 1) + c`

= `1/(n + 1)*(logx)^(n + 1) + c`