How to integrate this? (1 Viewer)

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[tex]\int{\frac{x^3+x^2+x-1}{x^2+2x+2}dx}[/tex]

I divided the polynomails and got:

[tex]\int{(x-1)dx}+\int{\frac{x+1}{x^2+2x+2}dx}[/tex]

This becomes:

[tex]x^2-x+\int{\frac{x}{x^+2x+2}dx}+\arctan{(x+2)}[/tex]

If I've done this right, how do I integrate the third term?

The arctan should have the (x+2) in brackets but I'm not too skilled in the use of Latex.
 
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Mr. Snookums said:
[tex]\int{\frac{x^3+x^2+x-1}{x^2+2x+2}dx}[/tex]

I divided the polynomails and got:

[tex]\int{(x-1)dx}+\int{\frac{x+1}{x^2+2x+2}dx}[/tex]
You can write the denominator as [itex](x+1)^2 + 1[/itex]. Thats one way out.
 
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[tex]\int{(x-1)dx}+\int{\frac{x+1}{x^2+2x+2}dx}[/tex]

You may consider trying a u substition. It may turn up producing something that looks familiar.
 
[tex]\int{\frac{x+1}{x^2+2x+2}dx}[/tex]

There is actually a simple way of integrating this expression. Try differentiating the denominator. What do you observe?
 

GCT

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There is actually a simple way of integrating this expression. Try differentiating the denominator. What do you observe?
Just change the denominator to (x+1)^2+1,

x+1=tan@

dx=sec^2 @ d@

You can then integrate

tan@sec^2@/(tan^2 @ +1) d@ which becomes

tan@sec^2@/sec^2 @ d@

thus you're left to integrate tan@ d@

I may have messed up on some trig identities here, it's been a while since I've worked with them.
 

Hootenanny

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I would recommend pizzasky's thoughts / process. Think about this integral;

[tex]\int\frac{f ' (x)}{f(x)} \;dx[/tex]

~H

Edit: Thank you Gokul, unfortunatly that is a mistake I make often, infact I made it today in a mock exam but I'm lucky they don't penalise it in Physics (they did in my maths exam :frown: )
 
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Gokul43201

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GCT, it can be done your way, but the observation pointed out by pizzasky makes the computation trivial (you have the answer in a single step).

Edit : Didn't read Hoot's post when I wrote this.
 
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GCT

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There is actually a simple way of integrating this expression. Try differentiating the denominator. What do you observe?
silly me, failed to catch on to that.
 
Ah, I see. Separating the second integral isn't needed because I can do it with substitution. Thank you.
 

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