How to Evaluate $\int\frac1{x \ln x+ 7 \ln x} \,\mathrm dx$

Question

I have tried many methods but do not know how to integrate this:

$$ \int \frac{1}{x\ln x + 7\ln x} dx $$

with respect to x.

Answer 1

$$\int\frac1{xln(x)+7ln(x)}dx$$ let $u=ln(x)$, then $ x=e^u du=\frac{dx}{x}=e^{-u}dx $
$$\int\frac{e^u}{e^uu+7u}du$$ Using integration by parts gives$$\frac{Ei(u)}{e^u+7}+\frac{e^uEi(u)}{e^{2u}+14e^u+49}=Li(x)\frac{x^2+15x+49}{x^2+14x+49}$$ which is not elementary.

Answer 2

I would factor the denominator into (ln x)(x+7). Then, you can split the fraction into (1/(x+7))*(1/ln x). From here, it is fairly easy to integrate by parts.

Answer 3

Let $u=\ln x$ ,

Then $x=e^u$

$dx=e^u~du$

$\therefore\int\dfrac{1}{x\ln x+7\ln x}dx=\int\dfrac{e^u}{u(e^u+7)}du$

Case $1$: $|7e^{-u}|\leq1$

Then $\int\dfrac{e^u}{u(e^u+7)}du$

$=\int\dfrac{1}{u(1+7e^{-u})}du$

$=\int\sum\limits_{n=0}^\infty\dfrac{(-1)^n7^ne^{-nu}}{u}du$

$=\ln u+\int\sum\limits_{n=1}^\infty(-1)^{n+1}7^nE_1(nu)+C$

$=\ln\ln x+\int\sum\limits_{n=1}^\infty(-1)^{n+1}7^nE_1(n\ln x)+C$

Case $2$: $|7e^{-u}|\geq1$

Then $\int\dfrac{e^u}{u(e^u+7)}du$

$=\int\dfrac{e^u}{7u\left(1+\dfrac{e^u}{7}\right)}du$

$=\int\sum\limits_{n=0}^\infty\dfrac{(-1)^ne^{(n+1)u}}{7^{n+1}u}du$

$=\sum\limits_{n=0}^\infty\dfrac{(-1)^n\text{Ei}((n+1)u)}{7^{n+1}}+C$

$=\sum\limits_{n=0}^\infty\dfrac{(-1)^n\text{Ei}((n+1)\ln x)}{7^{n+1}}+C$