Integrate $\int \frac {dx} {x^{27}\left( x^{7}-6\right) }$

Question

I tried to split it as $$\int\frac{1}{x^{34}}\frac{dx}{\left(1-\frac{6}{x^{7}}\right)}.$$

Okay so the question is incorrect but B is the correct answer as per the key , can anyone backderive

Answer 1

Observe that $$\frac{1}{x^{27}(x^7-a)} = x \cdot \frac{1}{u^4(u-a)},$$ where $u = x^7$, which facilitates a much easier decomposition via $u$: $$\frac{1}{x^{27}(x^7-a)} = \frac{x}{a^4(x^7-a)} - \frac{1}{a^4 x^6} - \frac{1}{a^3 x^{13}} - \frac{1}{a^2 x^{20}} - \frac{1}{a x^{27}}.$$ All but the first term have easy antiderivatives. The first term can be integrated by further decomposition, via the identity $$z^7 - 1 = (z-1)\prod_{k=1}^3 (z - \zeta_7^k)(z - \zeta_7^{-k}) = (z-1)\prod_{k=1}^3 \left( z^2 - 2 \cos \tfrac{2\pi k}{7} + 1 \right),$$ resulting in a product of linear and quadratic terms. I am not going to bother to complete the computation or integration: it is tedious and not illuminating.

---

Based on the picture you provided of the question, I have determined that there is a typographical error, and the integrand was intended to be $$\frac{1}{x^{22}(x^7-6)}.$$ It is only this particular exponent that will lead to an antiderivative that has anything remotely resembling the form that is claimed, for suitable constants or functions $A$, $B$, as then, the partial fraction decomposition will admit an obvious choice of substitution. If we backsolve by assuming $A$ is a scalar and $B = B(x)$ is a function of $x$, differentiating the RHS with respect to $x$ immediately yields $$\frac{1}{x^?(x^7-6)} = A B'(x)\left(\frac{1}{B(x)} - 18 + 18B(x) - 6B(x)^2\right).$$ Thus if the RHS is to be a rational function, $B(x)$ must itself be a rational function. If we substitute in the various choices, however, none fits. This is because the form of the antiderivative is itself flawed: the exponent in the integrand is not the only typographical error. Whomever wrote the question intended the answer to be (B), but the polynomial in $B$ is not correct.