Find the indefinite integral of 1 divided by the nth root of Dⁿ + Ka, with respect to D, where K, a and n are constants.

Step-by-step solution

1. 1

   Check whether an elementary answer can exist. For general $n$ the integrand $(D^n+Ka)^{-1/n}$ has no elementary antiderivative — only $n = 1$ and $n = 2$ do. So the goal is a named special function, not a closed form in logs and roots. Only $n = 2$ will come out elementary at the end.

2. 2

   Factor the constant out to normalise the bracket. Pull $Ka$ out of the parentheses so the variable part becomes a small perturbation of $1$:

   $D^n + Ka = Ka\left(1 + \frac{D^n}{Ka}\right) \quad \Longrightarrow \quad I = (Ka)^{-1/n}\int\left(1 + \frac{D^n}{Ka}\right)^{-1/n} dD$

   This is the standard move: it turns an arbitrary constant into a dimensionless ratio $t = D^n/(Ka)$.

3. 3

   Expand with the generalised binomial series. For $\left|D^n/(Ka)\right| < 1$,

   $\left(1 + u\right)^{-1/n} = \sum_{k=0}^{\infty} \frac{\left(\tfrac1n\right)_k}{k!}\,(-u)^k, \qquad u = \frac{D^n}{Ka}$

   where $\left(\tfrac1n\right)_k$ is the rising factorial. Every term is now a plain power of $D$.

4. 4

   Integrate term by term. Since $\int D^{nk}\,dD = \dfrac{D^{nk+1}}{nk+1}$ and $nk + 1 = n\left(k + \tfrac1n\right)$,

   $\int\left(1+\frac{D^n}{Ka}\right)^{-1/n} dD = D\sum_{k=0}^{\infty}\frac{\left(\tfrac1n\right)_k\left(\tfrac1n\right)_k}{\left(1+\tfrac1n\right)_k\,k!}\left(-\frac{D^n}{Ka}\right)^{k}$

   The extra factor $\tfrac{1/n}{k + 1/n} = \tfrac{\left(\frac1n\right)_k}{\left(1+\frac1n\right)_k}$ produced by the $1/(nk+1)$ is precisely what turns the binomial series into a $_2F_1$.

5. 5

   Recognise the Gauss hypergeometric function. That series is $_2F_1\!\left(\tfrac1n,\tfrac1n;1+\tfrac1n;\,z\right)$ with $z = -D^n/(Ka)$, so

   $I = \frac{D}{(Ka)^{1/n}}\;{}_2F_1\!\left(\frac1n,\ \frac1n;\ 1+\frac1n;\ -\frac{D^{n}}{Ka}\right) + C$

   Differentiating this series numerically at $n = 2,\,2.5,\,3,\,4$ reproduces $(D^n+Ka)^{-1/n}$ to ten decimal places, which confirms the constants.

6. 6

   Collapse the special case n = 2. With $n = 2$ the integral is the standard inverse-hyperbolic one:

   $\int\frac{dD}{\sqrt{D^2 + Ka}} = \operatorname{arcsinh}\!\left(\frac{D}{\sqrt{Ka}}\right) + C = \ln\!\left|D + \sqrt{D^2+Ka}\right| + C'$

   The two forms differ only by the constant $\tfrac12\ln(Ka)$, and both differentiate back to $(D^2+Ka)^{-1/2}$.