Question

Can someone explain to me what's happening in this integration?

Answer 1

Why do you solve this integration by using this method?\[\int\limits_{}^{}\frac{ x+1 }{ x^{2}(x+3) }\]

Answer 2

The method applied is:
\frac{x+1}{x^{2}(x+3) }= \frac{ax+b}{ x^{2} } + \frac{c}{x+3}

Answer 3

ooops. I'll write the method again

Answer 4

This is called partial fraction decomposition, it gets applied to polynomials that can be factored and have the form Grad\(f\) < Grad\(g\) where \(f\) and \(g\) are polynomials and your integral is in the form \(f/g\)

Answer 5

\[\frac{ x+1 }{ x ^{2}(x+3) }=\frac{ ax+b }{ x ^{2} }+\frac{ c }{ x+3 }\]

Answer 6

why do I have to do it like this???

Answer 7

The why I believe to already have answered, I assume you're asking for the why and not the how. I will try to be a bit more detail rich then.

Assume you're given the integral, as written in your problem. Obviously you cannot integrate it (nor do I). But the above equation tells you that after applying the method of partial fraction decomposition you can rewrite it as given in your equation. The result will be elementary integrals.
\[\large \int \frac{x+1}{x^2(x+3)}dx = \int \frac{ax+b}{x^2}dx + \int \frac{c}{x+3} \]
the left hand side is 'impossible' to integrate in that straight forward manner, the right hand side is elementary, but with partial fraction decomposition you have proved that both sides are identical, hence so are their integrals.

Answer 8

So when I solve this integration, I should leave the left side just like that and solve only the right side?

Answer 9

well of course you must first find \(a,b,c \in \mathbb{R}\) the numbers that describe the right hand side. But yes, you're right. You ignore the left side and integrate the right side, which is easy to integrate.

Answer 10

\[\frac{ (ax+b)(x+3) }{ x ^{2}(x+3) }+ \frac{ cx ^{2} }{ (x+3)x ^{2}}=\frac{ x+1 }{ x ^{2}(x+3) }\]

Answer 11

Don't you think we should solve it like this?

Answer 12

And the do it like this

Answer 13

\[\frac{ x ^{2}(a+c)+x(3a+b)+3b }{ x ^{2}(x+3) }\]

Answer 14

And put:

Answer 15

\[a+c=0, 3a+b=1, 3b=1\]

Answer 16

maybe, to be honest, your way is very very hard to understand. For instance
\[ \frac{ (ax+b)(x+3) }{ x ^{2}(x+3) }+ \frac{ cx ^{2} }{ (x+3)x ^{2}}=\frac{ x+1 }{ x ^{2}(x+3) } \tag{1}\]
What did you do here? this equation doesn't show up anywhere. you seem to have multiplied through by \(x^2(x+3)\) somehow, but if you did, why didn't you do so on the right hand side?
\[\frac{ x ^{2}(a+c)+x(3a+b)+3b }{ x ^{2}(x+3) } \tag{2} \]
(2) is an expression, not an equation anymore, it is missing the equation \(=\) relation. Although you might be right about your final conclusion that \(a+c=0, 3a+b=1, 3b=1\) there is no way to imply that from (2).

Answer 17

I didn't right the other parts cuz it would've took me much time xD Sorry

Answer 18

*write

Answer 19

The way you do it: You are given:
\[\frac{ x+1 }{ x ^{2}(x+3) }=\frac{ ax+b }{ x ^{2} }+\frac{ c }{ x+3 } \tag{*} \]
multiply * through by \(x^2(x+3)\) you will be left with:
\[\large x+1=(ax+b)(x+3)+cx^2 \]
now if you set \(x=-3\) you will be left with \(-2=9c\) if you set \(x=0\) you will be left with \(3b=1\) and if you set \(x=1\) you can find \(a\) because you know \(b \) and \(c\) now.

Answer 20

we seem to obtain the same results, so judging by that I will say you did it right :)

Answer 21

ok. Thank you :D

Answer 22

Do you know if there's a faster way to solve this?

Answer 23

the one above is one of the fastest I am aware of, because you only need to plugin values of \(x\) to make some terms vanish and you're given the answer immediately. But in general no, unfortunately partial fraction decomposition can be a very tedious exercise.

Answer 24

Ok, I understand. Thank you again for helping me out :D

Answer 25

you're very welcome