How do you find the integral from e^2 to e of $\frac{d x}{x \cdot ln (x^{8})}$ $dx / (x * ln (x^8))$ ?

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How do you find the integral from e^2 to e of $\frac{d x}{x \cdot ln (x^{8})}$ $dx / (x * ln (x^8))$ ?

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Answer 1 · 2017-02-01T19:21:42

$- \frac{1}{8} ln 2, or, - \frac{ln 2}{8} .$ $-1/8ln2, or, -ln2/8.$

Explanation:

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

$\frac{1}{(x) ({ln x}^{8})} = \frac{1}{x (8 ln x)} = \frac{1}{8 x ln x}$ $1/{(x)(lnx^8)}=1/{x(8lnx)}=1/(8xlnx)$

Hence, $I = \frac{1}{8} \int_{e^{2}}^{e} \frac{1}{x ln x} d x = \frac{1}{8} \int_{e^{2}}^{e} (\frac{1}{ln x}) (\frac{1}{x} d x)$ $I=1/8int_(e^2) ^e 1/(xlnx)dx=1/8int_(e^2) ^e (1/(lnx))(1/xdx)$

Substitute, $ln x = t \Rightarrow \frac{1}{x} d x = d t$ $ln x=t rArr 1/xdx=dt$

Also, $x = e^{2} \Rightarrow t = ln x = {ln e}^{2} = 2, &, x = e \Rightarrow t = 1 .$ $x=e^2 rArr t=ln x=ln e^2=2, &, x=e rArr t=1.$

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

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

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How do you find the integral from e^2 to e of $\frac{d x}{x \cdot ln (x^{8})}$ $dx / (x * ln (x^8))$ ?

1 Answer

Explanation:

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How do you find the integral from e^2 to e of $\frac{d x}{x \cdot ln (x^{8})}$ $dx / (x * ln (x^8))$ ?