Integral of a partial fraction function$\int\frac{1}{(x-1)^3(x-2)^2}dx$

Question

How do we determine the integral $$\int\frac{1}{(x-1)^3(x-2)^2}dx$$ My ideas are the followings:

1. We just split them into partial fractions like $$\frac{1}{(x-1)^3(x-2)^2}=\frac{A}{x-1}+\frac{B}{(x-1)^2}+\frac{C}{(x-1)^3}+\frac{D}{(x-2)}+\frac{E}{(x-2)^2}$$ But finding $A,B,C,D,E$ seems to be a really daunting task and the calculations will be huge.

2. I thought of substituting $x-1=u$ but that also doesn't help much since we again get the same form $\frac{1}{u^3(u-1)^2}$.

Is there any clever substitution to solve this? Trigonometric substitutions if possible are really helpful sometimes.

Answer 1

Clear the denominator to get $$1=A(x-1)^2(x-2)^2+B(x-1)(x-2)^2+C(x-2)^2+D(x-1)^3(x-2)+E(x-1)^3. \tag1$$ Substituting $x=1$ yields $1=C(1-2)^2$, so $C=1$. Substituting $x=2$ yields $1=E(2-1)^3$, so $E=1$. Substituting $x=0$ yields $$1=4A-4B+4C+2D-E=4A-4B+4+2D-1,$$ so $$4A-4B+2D=-2. \tag2$$ Substituting $x=3$ yields $$1 = 4A + 2B + C + 8D + 8E = 4A + 2B + 1 + 8D + 8,$$ so $$4A + 2B + 8D = -8. \tag3$$ Comparing coefficients of $x^4$ in $(1)$ yields $$0=A+D. \tag4$$ Now solve $(2)$ through $(4)$ to get $(A,B,D)=(3,2,-3)$.

Answer 2

Here a clever substitution $t=\frac{x-1}{x-2}$ to avoid the partial fraction $$ \int \frac{1}{(x-1)^3(x-2)^2}dx = \int \left(\frac1t-1\right)^3dt $$

Answer 3

For simplicity, letting $y=x-1$ transform the integral into $$\int \frac{1}{(x-1)^{3}(x-2)^{2}} d x=\int \frac{d y}{y^{3}(y-1)^{2}}$$

Repeatedly using $\frac{1}{y(y-1)}=\frac{1}{y-1}-\frac{1}{y}$ resolving the integrate into partial fractions as

$$ \begin{aligned} \frac{1}{y^{3}(y-1)^{2}} &=\frac{1}{y}\left[\frac{1}{y(y-1)}\right]^{2} \\ &=\frac{1}{y}\left(\frac{1}{y-1}-\frac{1}{y}\right)^{2} \\ &=\frac{1}{y(y-1)^{2}}-\frac{2}{y^{2}(y-1)}+\frac{1}{y^{3}}\\&=\frac{1}{y(y-1)} \cdot \frac{1}{y-1} -\frac{2}{y} \cdot \frac{1}{y(y-1)}+\frac{1}{y^{3}}\\&=\frac{1}{(y-1)^{2}}-\frac{1}{y-1}+\frac{1}{y}-\frac{2}{y-1}+\frac{2}{y}+\frac{2}{y^{2}}+\frac{1}{y^{3}}\\&= \frac{3}{x-1}+\frac{2}{(x-1)^{2}}+\frac{1}{(x-1)^{3}}-\frac{3}{x-2}+\frac{1}{(x-2)^{2}} \end{aligned} $$ We can now conclude that $$ \begin{aligned} I &=\int\left(\frac{3}{x-1}+\frac{2}{(x-1)^{2}}+\frac{1}{(x-1)^{3}}-\frac{3}{x-2}+\frac{1}{(x-2)^{2}}\right) d x \\ &=3 \ln |x-1|-\frac{2}{x-1}-\frac{1}{2(x-1)^{2}}-3 \ln |x-2|-\frac{1}{x-2}+C\\ &=\frac{-6 x^{2}+15 x-8}{2(x-2)(x-1)^{2}}+3 \ln \left|\frac{x-1}{x-2}\right|+C \end{aligned} $$

Answer 4

For simplicity, let $y=x-1$ $$ \frac{1}{y^{3}(y-1)^{2}}\equiv\frac{A}{y}+\frac{B}{y^{2}}+\frac{C}{y^{3}}+\frac{f(y)}{(y-1)^{2}}, $$ Then $$ 1\equiv A y^{2}(y-1)^{2}+B y(y-1)^{2}+C(y-1)^{2}+y^{3} f(y) \tag*{(1)} $$ Putting $y=0$ gives $C=1$ and rearranging $(1)$ yields $$ -y+2=A y(y-1)^{2}+B(y-1)^{2}+y^{2} f(y) \tag*{(2)} $$ Again, putting $y=0$ gives $B=2$ and rearranging $(2)$ yields $$ -2 y+3=A(y-1)^{2}+y f(y) \tag*{(3)} $$ Again, putting $y=0$ gives $A=3$ and rearranging $(3)$ yields

$$ -2 y+3-3(y-1)^{2}=y f(y) \Rightarrow f(y)=-3y+4 $$ Putting back gives us its partial fractions $$ \begin{aligned} \frac{1}{y^{3}(y-1)^{2}} &=\frac{3}{y}+\frac{2}{y^{2}}+\frac{1}{y^{3}}+\frac{-3 y+4}{(y-1)^{2}} \\ &=\frac{3}{y}+\frac{2}{y^{2}}+\frac{1}{y^{3}}-\frac{3}{y-1}+\frac{1}{(y-1)^{2}} \end{aligned} $$ $$ \begin{aligned} I &=\int\left(\frac{3}{x-1}+\frac{2}{(x-1)^{2}}+\frac{1}{(x-1)^{3}}-\frac{3}{x-2}+\frac{1}{(x-2)^{2}}\right) d x \\ &=3 \ln |x-1|-\frac{2}{x-1}-\frac{1}{2(x-1)^{2}}-3 \ln |x-2|-\frac{1}{x-2}+C\\ &=\frac{-6 x^{2}+15 x-8}{2(x-2)(x-1)^{2}}+3 \ln \left|\frac{x-1}{x-2}\right|+C \end{aligned} $$