Find the indefinite integral of 1 divided by the product of (x + 1) and (ln x) raised to the power a.

Step-by-step solution

1. 1

   Substitute u = ln x to put the two awkward pieces on the same footing. With $x = e^{u}$ and $dx = e^{u}\,du$,

   $\int \frac{dx}{(x+1)(\ln x)^{a}} = \int \frac{e^{u}}{\left(e^{u}+1\right)u^{a}}\,du = \int \frac{u^{-a}}{1+e^{-u}}\,du$

   using $\dfrac{e^{u}}{e^{u}+1} = \dfrac{1}{1+e^{-u}}$. The integrand is now a power times a Fermi-type factor, which is the recognisable non-elementary shape.

2. 2

   Expand the Fermi factor as a geometric series. For $u > 0$ (that is, $x > 1$) we have $e^{-u} < 1$, so

   $\frac{1}{1+e^{-u}} = \sum_{n=0}^{\infty} (-1)^{n} e^{-nu}$

   Splitting off the $n = 0$ term,

   $\int \frac{u^{-a}}{1+e^{-u}}\,du = \int u^{-a}\,du + \sum_{n=1}^{\infty}(-1)^{n}\int u^{-a}e^{-nu}\,du$

3. 3

   Integrate the leading term. For $a \ne 1$,

   $\int u^{-a}\,du = \frac{u^{1-a}}{1-a} = \frac{(\ln x)^{1-a}}{1-a}$

   and for the exceptional value $a = 1$ this term is $\ln|u| = \ln|\ln x|$ instead — the one case where the formula below must be replaced.

4. 4

   Express the remaining integrals with the upper incomplete gamma function. Since $\Gamma(s, z) = \int_{z}^{\infty} t^{s-1}e^{-t}\,dt$ satisfies $\dfrac{d}{du}\Gamma(s, nu) = -n^{s}u^{s-1}e^{-nu}$, setting $s = 1-a$ (so that $u^{s-1} = u^{-a}$) gives

   $\int u^{-a}e^{-nu}\,du = -n^{a-1}\,\Gamma(1-a,\, nu)$

5. 5

   Assemble the antiderivative. Substituting back $u = \ln x$ and absorbing the sign into the alternation:

   $\int \frac{dx}{(x+1)(\ln x)^{a}} = \frac{(\ln x)^{1-a}}{1-a} + \sum_{n=1}^{\infty}(-1)^{n+1} n^{a-1}\,\Gamma\!\left(1-a,\ n\ln x\right) + C$

   for $a \ne 1$ and $x > 1$.

6. 6

   Verify by differentiating the closed form. Term by term, $\frac{d}{dx}\Gamma(1-a, n\ln x) = -n^{1-a}(\ln x)^{-a}x^{-n-1}$, so the whole derivative collapses to a geometric sum:

   $\frac{(\ln x)^{-a}}{x}\left[1 + \sum_{n=1}^{\infty}\left(-\tfrac1x\right)^{n}\right] = \frac{(\ln x)^{-a}}{x}\cdot\frac{1}{1+\tfrac1x} = \frac{1}{(x+1)(\ln x)^{a}} \ \checkmark$

   As a second check, $a = 0$ must return the elementary $\ln(x+1)$: the formula gives $\ln x + \sum_{n\ge1}\frac{(-1)^{n+1}}{n}x^{-n} = \ln x + \ln\!\left(1+\tfrac1x\right) = \ln(x+1)$, confirmed numerically at $x = 1.5,\ 3,\ 7$.