Question

integrate -[lnxlnx]/x

Answer 1

Do you mean \[-\int\frac{\ln(x)^2}{x}dx\] ?

Answer 2

Or \[ -\int \frac{\ln(\ln(x)x)}{x} \] ?

Answer 3

i bet is second. use
\[u=\ln(x)\] and get the answer almost immediately

Answer 4

[lnx
]lnx

Answer 5

did u get what i mean?

Answer 6

\[-\int\limits_{}^{}\frac{\ln(x) \ln(x)}{x} dx =-\int\limits_{}^{}(\ln(x))^2 \cdot \frac{1}{x} dx =-\int\limits_{}^{} u^2 du\]
where u=ln(x) => du=dx/x