Question

`int(log(logx) + 1/(logx)^2)dx` = ______.

Options

• `xloglogx + x/logx + c`

• `xloglogx + (2x)/logx + c`

• `x^'loglogx - x/logx + c`

• `xloglogx - (2x)/logx + c`

Answer 1

`int(log(logx) + 1/(logx)^2)dx` = `underlinebb(x^'loglogx - x/logx + c)`.

Explanation:

I = `intx/x(log(logx) + 1/(logx)^2)dx` 

Let logx = t

⇒ `1/x dx` = dt

and x = e^t

I = `inte^t(logt + 1/t^2)dt`

= `inte^t(logt + 1/t)dt - inte^t(1/t - 1/t^2)dt`

= `e^tlogt - e^t. 1/t + c`

= `xlog(logx) - x/logx + c`