Integrating with Partial Fractions and Completing the Square: Tips and Tricks"

[Dan350]

I tried Partial fractions,, I completed the square which is (x-1)^2-3 ,, but,, what to do with the numer 3? i tried trig sub also,, didnt worked,,
Any ideas on how to integrate?
Is there a way to simplify??

Thanks a lot,,

 

[Curious3141]

I tried Partial fractions,, I completed the square which is (x-1)^2-3 ,, but,, what to do with the numer 3? i tried trig sub also,, didnt worked,,
Any ideas on how to integrate?
Is there a way to simplify??

Thanks a lot,,

Express the numerator as (2x-2) + 3.

Split up the expression. The first integrand is of the form f'(x)/f(x).

For the second integral, it's probably easiest to just factorise the expression, then use partial fractions. You may have some surds to deal with, but it's still quite simple.

Sorry for the multiple edits, I'm rushing for time and am doing this in my head.

 

[Dan350]

Express the numerator as (2x-2) + 3.

Split up the expression. The first integrand is of the form f'(x)/f(x).

Second integral can be done in one of two ways.

First is to complete the square as you've done, then use a hyperbolic trig sub (think cosh).

Second is to factorise the expression, then use partial fractions.

,, but why did you express it like that? (2x-2) + 3 thanks a lot, is see I can use use u sub since d/dx of x^2-2x-2 = 2x-2.. but i woul like to know how and why did you expressit like that,... thank you!

 

[Curious3141]

,, but why did you express it like that? (2x-2) + 3 thanks a lot, is see I can use use u sub since d/dx of x^2-2x-2 = 2x-2.. but i woul like to know how and why did you expressit like that,... thank you!

I hope you took notice of my edited post.

Why did I express it like that? It's just pattern recognition. I guess I just taught myself to recognise that form f'(x)/f(x) "hiding" in an integrand.

 

[Dan350]

I hope you took notice of my edited post.

Why did I express it like that? It's just pattern recognition. I guess I just taught myself to recognise that form f'(x)/f(x) "hiding" in an integrand.

? but the 3 , where did you get it from? Would you mind to explain me really quick? Thanks

 

[SteamKing]

(2x-2)+3 = ?

 

[Dan350]

(2x-2)+3 = ?

Yess
The procedure,, i would really appreciate

 

[Curious3141]

Yess
The procedure,, i would really appreciate

I figured out what f'(x) would be, then what I had to add on to it to make the numerator in the integrand.