1. ## Integration question

Integrate (x+3)dx / sqrt(x^2 - 2x + 5)

I got up to this and then my answer was:

Sqrt(x^2 - 2x + 5) + 4ln|(sqrt(x^2-2x+5) + x - 1) / 2| + C

But in the answer how did they randomly get rid of a half inside of the absolute value sign?

Sqrt(x^2 - 2x + 5) + 4ln|sqrt(x^2-2x+5) + x - 1| + C

So they just got rid of the half somehow

Thanks for any help

2. ## Re: Integration question

How did you get the 2nd part in ln? It should be in sine.

3. ## Re: Integration question

Originally Posted by pikachu975
Integrate (x+3)dx / sqrt(x^2 - 2x + 5)

I got up to this and then my answer was:

Sqrt(x^2 - 2x + 5) + 4ln|(sqrt(x^2-2x+5) + x - 1) / 2| + C

But in the answer how did they randomly get rid of a half inside of the absolute value sign?

Sqrt(x^2 - 2x + 5) + 4ln|sqrt(x^2-2x+5) + x - 1| + C

So they just got rid of the half somehow

Thanks for any help
Use log laws and so log|A/2| = log|A| – log(2). The – log(2) is just a constant so can be absorbed into the +C. The net effect is you can ignore the '/2' inside the log.

4. ## Re: Integration question

Hope this scrawl is readable and helpful - loads of stuff from the integration trick bag ... I didn't have room to re-sub back in each variable for the longer part...

Edit - sorry looks like you are well in control of all the method, I just read InteGrand response, I see the confusion. Oh well that was fun!

5. ## Re: Integration question

Originally Posted by si2136
How did you get the 2nd part in ln? It should be in sine.
I think I thought that at first but the square root has plus not minus so I just did trig substitution

Originally Posted by InteGrand
Use log laws and so log|A/2| = log|A| – log(2). The – log(2) is just a constant so can be absorbed into the +C. The net effect is you can ignore the '/2' inside the log.
Thanks, didn't realise this.

Originally Posted by weaknuclearforce
Hope this scrawl is readable and helpful - loads of stuff from the integration trick bag ... I didn't have room to re-sub back in each variable for the longer part...

Edit - sorry looks like you are well in control of all the method, I just read InteGrand response, I see the confusion. Oh well that was fun!
Thanks for the help anyway, but just a note to integrate secx just multiply top and bottom by (secx+tanx)

6. ## Re: Integration question

Of course! You can tell I didn't bother following through with t-sub haha - is this a typical Ext2 Q, seems a little on the 'lets use every 1st yr Uni trick we can' - poor HSC kids, I don't envy you!

7. ## Re: Integration question

Originally Posted by weaknuclearforce
Of course! You can tell I didn't bother following through with t-sub haha - is this a typical Ext2 Q, seems a little on the 'lets use every 1st yr Uni trick we can' - poor HSC kids, I don't envy you!
It's a classic trick to integrate sec(x) (the multiply top and bottom by sec + tan).

The question in general is mainly just slightly tedious calculations I think. The method for doing it is standard and expected to be known by 4U students. (Write numerator as a*q'(x) + b for some wisely chosen constants a and b, and go from there, recalling the technique of completing the square for integrating something like 1/(sqrt(q(x)). Here, q(x) represents the quadratic in the radical and q'(x) its derivative.)

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