Partial fraction method in Integration

Question

My question is can we proceed with the method without even factorising the denominator. For eg take a question. $$\int \frac{2x-3}{(x^2-1)(2x+3)}dx$$ Can we write directly $$\frac{2x-3}{(x^2-1)(2x+3)}=\frac{Ax+B}{x^2-1}+\frac{C}{2x-3}$$ without simplifying denominator term $(x^2-1)=(x-1)(x+1)$?

Answer 1

There is no problem in this; it's just a matter of choice. HOWEVER, if you resolve the integrand as $$\frac{2x-3}{(x^2-1)(2x+3)}=\frac{Ax+B}{x^2-1}+\frac{C}{2x-3}$$ then how would you go about integrating $(Ax+B)/(x^2-1)$?

Unless you are lucky that $B=0$, (in the above case you get $B=1$), in all other cases, you will have no choice but again resolving it into two more fractions, so why not do it directly?

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Edit

Suppose we get $A=5$ and $B=0$ in the above one, then integrating is trivial. $$\int \frac{5x}{x^2-1}\ dx=\int\frac 5 2 \frac{2x}{x^2-1}\ dx$$ $$ = \frac{5}{2}\ln |x^2-1|$$ However, say we get $B=1$, then how would you solve $$\int\frac{5x+1}{x^2-1}$$ You have no other choice but to use partial fraction decomposition.

Hope this is clear :)