# How to integrate this? (1 Viewer)

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#### Mr. Snookums

$$\int{\frac{x^3+x^2+x-1}{x^2+2x+2}dx}$$

I divided the polynomails and got:

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

This becomes:

$$x^2-x+\int{\frac{x}{x^+2x+2}dx}+\arctan{(x+2)}$$

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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#### maverick280857

Mr. Snookums said:
$$\int{\frac{x^3+x^2+x-1}{x^2+2x+2}dx}$$

I divided the polynomails and got:

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

#### Hammie

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

You may consider trying a u substition. It may turn up producing something that looks familiar.

$$\int{\frac{x+1}{x^2+2x+2}dx}$$

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

#### GCT

Homework Helper
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

Staff Emeritus
Gold Member

$$\int\frac{f ' (x)}{f(x)} \;dx$$

~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 )

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#### Gokul43201

Staff Emeritus
Gold Member
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

Homework Helper
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.

#### Mr. Snookums

Ah, I see. Separating the second integral isn't needed because I can do it with substitution. Thank you.

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