Q:

Can you solve this integral without using partial fraction?

It can be done by using partial fraction but it becomes complex. Is it possible to solve this in any other way?!
$$
\int \frac{dx}{(x-1)^2(x-2)^3}
$$

A:

Hint:
Write the numerator as $$((x+2)-(x+1))^3$$
Then $x+1=x+2-1$ and $x+2=x+1+1$
Alternatively use integration by parts

A:

We may avoid partial fraction with a little trick.
$$\int \frac{dx}{(x-1)^2(x-2)^3}=\int \frac{(x-1)dx}{(x-1)^3(x-2)^3} $$
Note that $$x-1 = (x-3/2) +1/2 = u+1/2$$ while $$x-2 = (x-3/2) -1/2 =u-1/2$$ where $ u=(x-3/2).$
Substitute in the integral  to get $$ \int \frac{(x-1)dx}{(x-1)^3(x-2)^3} =\int \frac{(u+1/2)du}{(u^2-1/4)^3}=\int \frac{udu}{(u^2-1/4)^3}+\int \frac{(1/2)du}{(u^2-1/4)^3} $$ 
These integrals do not require partial fractions.

