How to Integrate Polynomials and Solve for Unknown Terms?

[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.

 

[maverick280857]

\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.

 

[pizzasky]

\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]

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]

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

\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 )

 

[Gokul43201]

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.

 

[GCT]

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.