How do you find the integral from e^2 to e of #dx / (x * ln (x^8))#?

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Answer 1 · 2017-02-01T19:21:42

#-1/8ln2, or, -ln2/8.#

Using the usual rules of Log function, we find that the Integrand is

#1/{(x)(lnx^8)}=1/{x(8lnx)}=1/(8xlnx)#

Hence, #I=1/8int_(e^2) ^e 1/(xlnx)dx=1/8int_(e^2) ^e (1/(lnx))(1/xdx)#

Substitute, #ln x=t rArr 1/xdx=dt#

Also, #x=e^2 rArr t=ln x=ln e^2=2, &, x=e rArr t=1.#

#:. I=1/8 int_2^1 1/t dt=1/8[ln|t|]_2^1=1/8(ln1-ln2)#

#"Therefore, I="-ln2/8.#

Enjoy Maths.!