Question

(log x² - log x). log 1/x+ (log x)² = 0 ​

Answer 1

Answer:

Taking (logx)

2

as first function and 1 as second function and integrating by part, we obtain

I=(logx)

2

∫xdx−∫[{(

dx

d

logx)

2

}∫xdx]dx

=

2

x

2

(logx)

2

−[∫2logx⋅

x

1

⋅

2

x

2

dx]

=

2

x

2

(logx)

2

−∫xlogxdx

Again integrating by parts, we obtain

I=

2

x

2

(logx)

2

−[logx∫xdx−∫{(

dx

d

logx)∫xdx}dx]

=

2

x

2

(logx)

2

−[

2

x

2

−logx−∫

x

1

⋅

x

x

2

dx]

=

2

x

2

(logx)

2

−

2

x

2

logx+

2

1

∫xdx

=

2

x

2

(logx)

2

−

2

x

2

logx+

4

x

2

+C