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**Hi I'm currently doing 'integral by substitution' part in a book.**

Although it is integral by substitution part, some exercises are solved using reduction of fraction and integral, without substitution.

(Actually I can't solve some exercises if I use substitution and the book's explanation also doesn't use substitution)

By the way I'm stuck with this exercise:

##\int \frac{2x+6}{(x-1)(x+1)^2} dx##

I tried to use substitution but that makes the problem more complicated and I can't also use reduction of fraction.

So I looked up the book's explanation, the book solved this problem in a weird way :

##

\frac{2x+6}{(x-1)(x+1)^2}

= \frac{a}{x-1} + \frac{b}{x+1} + \frac{c}{(x+1)^2}

=\frac{a(x+1)^2+b(x-1)(x+1)+c(x-1)}{(x-1)(x+1)^2}

##

##

2x+6 = (a+b)x^2+(2a+c)x+a-b-c

##

##

a+b=0, 2a+c=2, a-b-c=6

##

##

a=2,b=-2,c=-2

##

##

\int \frac{2x+6}{(x-1)(x+1)^2} dx = \int (\frac{2}{x-1}-\frac{2}{x+1}-\frac{2}{(x+1)^2})dx

##

##

=2\ln|x-1|-2\ln|x+1|+\frac{2}{x+1}+C

##

The book didn't introduced technique like this.

And I feel unnatural and not intuitive about its process because..

Who would imagine to split the expression

##

\frac{2x+6}{(x-1)(x+1)^2}

##

into

##

\frac{a}{x-1} + \frac{b}{x+1} + \frac{c}{(x+1)^2}

##

I can't even use only a, b instead of a, b, c:

##

\frac{2x+6}{(x-1)(x+1)^2} = \frac{a}{x-1}+\frac{b}{x+1}=\frac{a(x+1)^2+b(x^2-1)}{(x-1)(x+1)^2}=\frac{(a+b)x^2+2ax+a-b}{(x-1)(x+1)^2}

##

##

a+b=0, 2a=2,a-b=6

##

And that is contradiction 'a' should be 1 because 2a=2

and b should be -5 because a-b=6

but that doesn't make a+b=0

I don't know why this doesn't work.

So my question is

1. why last thing (use only a, b instead of a, b, c) doesn't work?

2. How can I think this way immediately when I see an exercise like this? Is this technique frequently used?? and how can I think to split the expression into exactly a,b,c, not a, b

Thanks for reading.split the expression

##

\frac{2x+6}{(x-1)(x+1)^2}

##

into

##

\frac{a}{x-1} + \frac{b}{x+1} + \frac{c}{(x+1)^2}

##