`I=int \ log_e (log_ex)/(x(log_e x))dx`

I=int(log_e( log_ex))/(x(log_ex))dx is equal to

The integral inte^x(f(x)+f\'(x))dx can be solved by using integration by parts such that: I=inte^xf(x)dx+inte^xf\'(x)dx=e^xf(x)-inte^xf\'(x)dx+inte^xf\'(x)dx=e^xf(x)+C , and inte^(ax)(f(x)+(f\'(x))/a)dx=e^(ax)f(x)/a+C ,Now answer the question: int{log_e(log_ex)+1/(log_ex)^2}dx is equal to (A) log_e(log_ex)+C (B) xlog_e(log_ex)-x/log_ex+C (C) x/log_ex-log_ex+C (D) log_e(log_ex)-x/log_ex+C

The integral inte^x(f(x)+f\'(x))dx can be solved by using integration by parts such that: I=inte^xf(x)dx+inte^xf\'(x)dx=e^xf(x)-inte^xf\'(x)dx+inte^xf\'(x)dx=e^xf(x)+C , and inte^(ax)(f(x)+(f\'(x))/a)dx=e^(ax)f(x)/a+C ,Now answer the question: int{log_e(log_ex)+1/(log_ex)^2}dx is equal to (A) log_e(log_ex)+C (B) xlog_e(log_ex)-x/log_ex+C (C) x/log_ex-log_ex+C (D) log_e(log_ex)-x/log_ex+C

int log_(e)xdx=int(1)/(log_(x)e)dx=

int(e^(log_(e)x))/(x)dx

int(e^(log_(e)x))/(x)dx

int(e^(log_(e)x))/(x)dx

int e^(a log_(e)x)dx

int x log_(e)(1+x)dx

Evaluate: int(e^(5(log)_e x)-e^(4(log)_ex))/(e^(3(log)_e x)-e^(2logx))dx