a/bc = a/b - a/c?

[Al-Layth]

i just saw somebody decompose the integrand
$$\int_{0}^{\infty}{\frac{\sin^2(x)}{x^2(x^2+1)}}dx = \int_{0}^{\infty } \frac{\sin^2(x)}{x^2}dx - \int_{0}^{\infty } \frac{\sin^2(x)}{x^2+1}dx$$

is this okay in general or specific to improper integrals???

 

[Dr.Peterson]

i just saw somebody decompose the integrand
$$\int_{0}^{\infty}{\frac{\sin^2(x)}{x^2(x^2+1)}}dx = \int_{0}^{\infty } \frac{\sin^2(x)}{x^2}dx - \int_{0}^{\infty } \frac{\sin^2(x)}{x^2+1}dx$$

is this okay in general or specific to improper integrals???

Why do you think it is wrong in general that

?
Do you not recognize this as partial fraction decomposition?

(It is not true in general, of course, that a/bc = a/b - a/c. It is important here that c-b=1.)

What specific issue are you concerned about? (I haven't tried to carry out the integrals yet.)

 

[NotJeffM]

$$\text {If } \exists \ A \text { and } B \text { such that } \dfrac{A}{x^2} + \dfrac{B}{x^2 + 1} = \dfrac{\sin^2(x)}{x^2(x^2 + 1},\\ \text {then } \dfrac{A(x^2 + 1) + Bx^2}{x^2(x^2 + 1)} = \dfrac{\sin^2(x)}{x^2(x^2 + 1)} \implies \\ A(x^2 + 1) + Bx^2 = \sin^2 (x) \implies x^2(A + B) + A = x^2 * 0 + \sin^2(x) \implies \\ A + B = 0 \text { and } A = \sin^2(x) \implies B = - A \implies B = - \sin^2(x).\\ \text {Check.}\\ \dfrac{\sin^2(x)}{x^2} - \dfrac{\sin^2(x)}{x^2 + 1} = \dfrac{\sin^2(x) * (x^2 + 1) - \sin^2(x) * x^2}{x^2(x^2 + 1)} =\\ \dfrac{\cancel {\sin^2(x) * x^2} + \sin^2(x) * 1 - \cancel {\sin^2(x) * x^2}}{x^2(x^2 + 1)} = \dfrac{\sin^2(x)}{x^2(x^2 + 1)}. \ \checkmark$$
The result is specific to this fraction. The method is quite general, and it relates to fractions rather than just integrals.