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

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

1.	`int{log(logx)+(1)/((logx)^(2))}dx=x {f (x)-g(x)}+C`, thenA. `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)`
Answer» Correct Answer - 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)`