Evaluating integral of the form $\int \frac {dx}{(x-b)^m(x-a)^n}$

Question

$$\int \frac {dx}{(x-b)^3(x-a)^2}$$

Can someone please help evaluate this integral? Finding the partial fraction seems tedious and I don't know any other way.

Answer 1

Hint: Rewrite the $1$ on the numerator as $1 = \dfrac{(x-a) - (x-b)}{b-a}$,and split it into $2$ fractions, and repeat this trick until the fraction becomes simple enough to have a clear antiderivative.

Answer 2

Let we assume $\displaystyle I =\int\frac{1}{(x-a)^2(x-b)^3}dx$

Then put $\displaystyle (x-a)=t(x-b)\Longrightarrow t=\frac{x-a}{x-b}$

So $\displaystyle t=\frac{(x-b)+(b-a)}{x-b}=1+\frac{b-a}{x-b}$

So we have $\displaystyle x-b=\frac{b-a}{t-1}$ and $\displaystyle dx=\frac{a-b}{(t-1)^2}dt$

So we have $\displaystyle I=\frac{1}{(a-b)^4}\int\frac{(t-1)^3}{t^2}dt$

Answer 3

Trick: find $$ F(a,b,x) = \int \frac{1}{(x-a)(x-b)}\,dx $$ which is easy, since $\frac{1}{(x-a)(x-b)}=\frac{1}{a-b}\left(\frac{1}{x-a}-\frac{1}{x-b}\right) $, then differentiate once with respect to $a$ and twice with respect to $b$.

Answer 4

Case where $n,m$ are not integers?
We want to do $$ I(a,b,n,m;x) = \int\frac{dx}{(x-a)^n(x-b)^m} \tag1$$ where $a < b$ and $m,n$ are real numbers. Say we want the antiderivative where $x>b$ so that those fractional powers are real.

First, do the special case $$ I(0,1,n,m;y) = \int\frac{dy}{y^n(y-1)^m} = \frac{y^{1-n-m}}{1-n-m}\;{}_2F_1\left(m,m+n-1;n+m;\frac{1}{y}\right) \tag2$$ then change variables $y=\frac{x-a}{b-a}$ in $(1)$ to get: $$ I(a,b,m,n,x) = \int\frac{dx}{(x-a)^n(x-b)^m} = (b-a)^{1-n-m}\int \frac{dy}{y^n(y-1)^m} \\ = (b-a)^{1-n-m}I(0,1,n,m,y) = (b-a)^{1-n-m}I\left(0,1,n,m,\frac{x-a}{b-a}\right) \\ = \frac{(x-a)^{1-n-m}}{1-n-m}\;{}_2F_1\left(m,m+n-1;n+m;\frac{b-a}{x-a}\right) $$

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In case $n+m=1$, we get $$ I(0,1,n,1-n,y) = (n-1)\;{}_3F_2\left(1,1,2-n;2,2;\frac{1}{y}\right) -\gamma+\log n-\psi(n) . $$