Computing $int_{alpha}frac{1}{(z-a)^n(z-b)^m}dz$ using Cauchy integral formula

Answering my own question, thanks to the comments. Let $f(z) = frac{1}{(z-b)^m}$, then by Cauchy integral formula $$int_{alpha}frac{1}{(z-a)^n(z-b)^m}dz = int_{alpha}frac{f(z)}{(z-a)^n}dz = frac{2pi i}{(n-1)!}f^{(n-1)}(a).$$ Trying to find an expression of $f^{(n-1)}(z)$ begin{align*} f^{(1)} &= -mfrac{1}{(z-b)^{m+1}}\ f^{(2)} &= m(m+1)frac{1}{(z-b)^{m+2}}\ f^{(3)} &= -m(m+1)(m+2)frac{1}{(z-b)^{m+3}}\ &vdots\ f^{(n-1)} &= (-1)^{n-1}m(m+1)(m+2)dotsc(m+n - 2)frac{1}{(a-b)^{m+n-1}} end{align*} So, begin{align*} int_{alpha}frac{1}{(z-a)^n(z-b)^m}dz &= frac{2pi i}{(n-1)!}(-1)^{n-1}m(m+1)(m+2)dotsc(m+n - 2)frac{1}{(a-b)^{m+n-1}}\ &= frac{2pi i}{(n-1)!}(-1)^{n-1-m-n+1}m(m+1)(m+2)dotsc(m+n - 2)frac{1}{(b-a)^{m+n-1}}\ &=frac{2pi i}{(n-1)!}(-1)^{-m}m(m+1)(m+2)dotsc(m+n - 2)frac{1}{(b-a)^{m+n-1}}\ &=frac{2pi i}{(n-1)!}(-1)^{m}m(m+1)(m+2)dotsc(m+n - 2)frac{1}{(b-a)^{m+n-1}}. end{align*} Note that $${n + m -2 choose n-1} = frac{(n+m-2)(n+m-3)dotsc m}{(n-1)!}.$$ Hence, $$int_{alpha}frac{1}{(z-a)^n(z-b)^m}dz = 2pi i(-1)^m{n + m -2 choose n-1}frac{1}{(b-a)^{m+n-1}}.$$

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begin{align} &bbox[5px,#ffd]{left. oint_{verts{z} = r},, {dd z over pars{z - a}^{,n}pars{z - b}^{,m}} ,rightvert_{substack{n, m in mathbb{N} \[1mm] verts{a} < r < verts{b}}}} \[5mm] = & 2piicon{Res}bracks{{1 over pars{z - a}^{,n} pars{z - b}^{,m}}, z = a}label{1}tag{1} end{align}

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begin{align} mbox{Note that} & {1 over pars{z - b}^{m}} = {1 over bracks{a - b + pars{z - a}}^{m}} \[5mm] & = {1 over pars{a - b}^{m}} bracks{1 + {z - a over a - b}}^{-m} \[5mm] & = {1 over pars{a - b}^{m}} sum_{k = 0}^{infty}{-m choose k} pars{z - a over a - b}^{k} end{align} (ref{1}) becomes begin{align} &bbox[5px,#ffd]{left. oint_{verts{z} = r},, {dd z over pars{z - a}^{,n}pars{z - b}^{,m}} ,rightvert_{substack{n, m in mathbb{N} \[1mm] verts{a} < r < verts{b}}}} \[5mm] = & 2piic{1 over pars{a - b}^{m}} bracks{{-m choose n - 1} pars{1 over a - b}^{n - 1}} \[5mm] = & 2piic{1 over pars{a - b}^{m + n - 1}} {m + bracks{n - 1} - 1 choose n - 1}pars{-1}^{n - 1} \[5mm] = & bbx{2piic,pars{-1}^{m} {n + m - 2 choose n - 1} {1 over pars{b - a}^{n + m - 1}}} \ & end{align}