I want to know how does the following equation is true (I have checked for different values in Wolfram Alpha and the answer obtained from both sides are equal) $$\int_u^\infty \left[1-\frac{1}{\left(1+\frac{a}{x^k}\right)^m}\right] x \, dx = -\frac{u^2}{2} + \frac{u^2}{2} \left[_2F_1(m;-\frac{2}{k};1-\frac{2}{k};\frac{a}{u^k})\right]$$where $a,m,u$ are all positive values and $k>2$.
My attempt:
In my attempt I can use $v=\left(\frac{u}{x}\right)^k$ I can find $$\text{when } x=u \to v=1 \text{ and when } x=\infty \to v=0$$ $$\frac{dx}{dz}=-\frac{u}{k}v^{-\frac{1}{k}-1}$$ therefore with substitution we can have $$\int_u^\infty\left[1-\frac{1}{\left(1+\frac{a}{x^k} \right)^m} \right] x \, dx=\int_0^1\left[1-\frac{1}{\left(1+\frac{av}{u^k}\right)^m}\right]uv^{-\frac{1}{k}}\frac{u}{k}v^{-\frac{1}{k}-1}dv$$ $$\int_u^\infty\left[1-\frac{1}{\left(1+\frac{a}{x^k} \right)^m} \right] x \, dx = \int_0^1 \left[1-\left(\frac{1}{1+\frac{av}{u^k}}\right)^m\right](\frac{u^2}{k}) v^{-\frac{2}{k}-1} \,dv$$ now if somehow there is $v^{\frac{2}{k}-1}$ instead of $v^{-\frac{2}{k}-1}$ on right side of my last equation then the expression on the right side of my first equation can be obtained (using Eq 3.194 of Gradeshteyn book (also given below)). However, I do not know where I am making a mistake and how to show that the relation in my first equation is true analytically. Please help me in getting it right. Many many thanks in advance.
Edit: After the steps of Harry Peter
I am very thankful to Harry Peter for his help (see the Hint in the answer below). Here, I extend the steps of Harry Peter further. The last step of Harry Peter is $$\frac{1}{ak}\int_1^{\frac{a}{u^k}+1}\left(\frac{x}{a}-1\right)^{-\frac{2}{k}-1}(1-x^{-m})dx$$ now we can write it as $$\frac{1}{ak}\int_1^{\frac{a}{u^k}+1}\left(\frac{x}{a}-1\right)^{-\frac{2}{k}-1}dx-\frac{1}{ak}\int_1^{\frac{a}{u^k}+1}\left(\frac{x}{a}-1\right)^{-\frac{2}{k}-1}x^{-m}dx$$ which can be further written as $$\frac{1}{ak}\left(\frac{-k}{2}\right)(a)\left[\frac{x}{a}-1\right]^{-\frac{2}{k}}\bigg{|}_1^{\frac{a}{u^k}+1}-\frac{(-1)^{-\frac{2}{k}-1}}{ak}\int_1^{\frac{a}{u^k}+1}\left(1-\frac{x}{a}\right)x^{-m}dx$$ putting limits in the first part and using substitution $z=\frac{x}{a}$ in the second part we can write $$=-\frac{1}{2}\left[\left(\frac{1}{u^k}+\frac{1}{a}-1\right)^{-\frac{2}{k}}-\left(\frac{1}{a}-1\right)^{-\frac{2}{k}}\right]-\frac{1}{ak}(-1)^{-\frac{2}{k}-1}\int_{\frac{1}{a}}^{\frac{1}{u^k}+\frac{1}{a}}(1-z)^{-\frac{2}{k}-1}(az)^{-m}a \, dz$$ which after simplification can be written as $$=-\frac{1}{2}\left[\left(\frac{1}{u^k}+\frac{1}{a}-1\right)^{-\frac{2}{k}}-\left(\frac{1}{a}-1\right)^{-\frac{2}{k}}\right]-(-1)^{-\frac{2}{k}-1}a^{-m}\int_{\frac{1}{a}}^{\frac{1}{u^k}+\frac{1}{a}}(1-z)^{-\frac{2}{k}-1}z^{-m}\,dz$$ $$\frac{1}{ak}\int_1^{\frac{a}{u^k}+1}\left(\frac{x}{a}-1\right)^{-\frac{2}{k}-1}(1-x^{-m})dx=-\frac{1}{2}\left[\left(\frac{1}{u^k}+\frac{1}{a}-1\right)^{-\frac{2}{k}}-\left(\frac{1}{a}-1\right)^{\frac{2}{k}}\right]-C$$ where $C$ is given as follows $$C=(-1)^{-\frac{2}{k}-1}a^{-m}\int_{\frac{1}{a}}^{\frac{1}{u^k}+\frac{1}{a}}(1-z)^{-\frac{2}{k}-1}z^{-m}\,dz$$ $$C=(-1)^{-\frac{2}{k}-1}a^{-m}\left[\int_0^{\frac{1}{u^k}+\frac{1}{a}}(1-z)^{-\frac{2}{k}-1}z^{-m}\,dz-\int_0^{\frac{1}{a}}(1-z)^{-\frac{2}{k}-1}z^{-m}\,dz\right]$$ Now using the definition of incomplete Beta function ($B_z(p,q)=\int_0^zt^{p-1}(1-t)^{q-1}\,dt$) we can write $$C=(-1)^{-\frac{2}{k}-1}a^{-m}\left[B_{\frac{1}{u^k}+\frac{1}{a}}(1-m,-\frac{2}{k})-B_{\frac{1}{a}}(1-m,-\frac{2}{k})\right]$$ Now using the definition of Hypergeometric function ($B_z(p,q)=\frac{z^p}{p}\, _2F_1(p,1-q;p+1;z)$)we can write $$C=(-1)^{-\frac{2}{k}-1}a^{-m}\left[\frac{\left(\frac{1}{u^k}+\frac{1}{a}\right)^{1-m}}{1-m}\, _2F_1(1-m,1+\frac{2}{k};2-m;\frac{1}{u^k}+\frac{1}{a})-\frac{\left(\frac{1}{a}\right)^{1-m}}{1-m}\, _2F_1(1-m,1+\frac{2}{k};2-m;\frac{1}{a})\right]$$ Putting the value of $C$ in the above equation for $\frac{1}{ak}\int_1^{\frac{a}{u^k}+1}\left(\frac{x}{a}-1\right)^{-\frac{2}{k}-1}(1-x^{-m})dx$ we can write $$\frac{1}{ak}\int_1^{\frac{a}{u^k}+1}\left(\frac{x}{a}-1\right)^{-\frac{2}{k}-1}(1-x^{-m})dx=-\frac{1}{2}\left[\left(\frac{1}{u^k}+\frac{1}{a}-1\right)^{-\frac{2}{k}}-\left(\frac{1}{a}-1\right)^{-\frac{2}{k}}\right]-(-1)^{-\frac{2}{k}-1}a^{-m}\left[\frac{\left(\frac{1}{u^k}+\frac{1}{a}\right)^{1-m}}{1-m}\, _2F_1(1-m,1+\frac{2}{k};2-m;\frac{1}{u^k}+\frac{1}{a})-\frac{\left(\frac{1}{a}\right)^{1-m}}{1-m}\, _2F_1(1-m,1+\frac{2}{k};2-m;\frac{1}{a})\right]$$ Now the right side is not looking the same as the right side of my very first equation of this post. Maybe there is some very nice property or identity of Hypergeometric function that can be used to make it look like the right side of very first equation. But unfortunately I am unaware of any such formula. I request you to help me in getting the right side of very first equation. Many many thanks in advance.
Formula from Gradeshteyn book
Thank you
Hint: \begin{equation} \int x\left(1+\frac{a}{x^{k}}\right)^{-m} \mathrm{d}x \end{equation}
Let $y = -\frac{a}{x^{k}}$ \begin{align} \int x\left(1+\frac{a}{x^{k}}\right)^{-m} \mathrm{d}x &= \frac{1}{k} (-1)^{1+\frac{2}{k}} a^{\frac{2}{k}} \int (1-y)^{-m} y^{-1-\frac{2}{k}} \mathrm{d}y \\ &= \frac{1}{k} (-1)^{1+\frac{2}{k}} a^{\frac{2}{k}} \mathrm{B}_{y} \left(-\frac{2}{k},1-m \right) \\ &= \frac{1}{k} (-1)^{1+\frac{2}{k}} a^{\frac{2}{k}} y^{-\frac{2}{k}} (-1) \frac{k}{2} \,{}_{2}\mathrm{F}_{1} \left(-\frac{2}{k},m;1-\frac{2}{k};y \right) \\ &= \frac{x^{2}}{2} \,{}_{2}\mathrm{F}_{1} \left(-\frac{2}{k},m;1-\frac{2}{k};-\frac{a}{x^{k}} \right) \end{align}
Notes: 1. \begin{equation} \mathrm{B}_{z}(p,q) = \int_{0}^{z} t^{p-1} (1-t)^{q-1} \mathrm{d}t \end{equation} is the incomplete beta function.
2. \begin{equation} \mathrm{B}_{z}(p,q) = \frac{z^{p}}{p} {}_{2}\mathrm{F}_{1}(p,1-q;p+1;z) \end{equation} is the incomplete beta function in terms of Gauss's hypergeometric function.