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int{log(logx)+(1)/((logx)^(2))}dx=x {f (...

`int{log(logx)+(1)/((logx)^(2))}dx=x {f (x)-g(x)}+C`, then

A

`f(x)=log(logx),g(x)=(1)/(logx)`

B

`f(x)=logx, g(x)=(1)/(logx)`

C

`f(x)=(1)/(logx),g(x)=log(logx)`

D

`f(x)=(1)/(xlogx),g(x)=(1)/(logx)`

Text Solution

Verified by Experts

The correct Answer is:
A
Given, `int[log(logx)+(1)/((logx)^(2))]dx`
`=x[f(x)-g(x)]+C`
`LHS=int underset("II")(1).log underset("I")((logx))dx+int(1)/((logx)^(2))dx`
On integration by parts, we get
`xlog(logx)-int(1)/(logx)dx+int(1)/((logx)^(2))dx `
Again using integration by parts, we get
`xlog(logx)-(x)/(logx)-int(1)/((logx)^(2))dx+int(1)/((logx)^(2)dx+C`
`=x[log(logx)-(1)/(logx)]+C`
`therefore f(x)=log(logx),g(x)=(1)/(logx)`
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