Question

integration [log(logx^2)+(logx)^-2]dx​

Answer 1

Answer:

∫(log(logx)+

(logx)

2

1

)dx

=∫loglogxdx+∫

(logx)

2

1

dx

Integrating by parts, we get

Let u=log(logx)⇒du=

xlogx

1

dx

dv=dx⇒v=x

=xlog(logx)−∫x×

xlogx

1

dx+∫

(logx)

2

1

dx

=xlog(logx)−∫

logx

dx

+∫

(logx)

2

1

dx

Let u=(logx)

−1

⇒du=−

x(logx)

2

1

dx

dv=dx⇒v=x

=xlog(logx)−[

logx

x

−∫x×−

x(logx)

2

1

]+∫

(logx)

2

1

dx

=xlog(logx)−[

logx

x

+∫

(logx)

2

dx

]+∫

(logx)

2

1

dx

=xlog(logx)−

logx

x

−∫

(logx)

2

dx

+∫

(logx)

2

1

dx+c

=xlog(logx)−

logx

x

+c