How did you get the 2nd part in ln? It should be in sine.
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?
The answer was basically
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
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How did you get the 2nd part in ln? It should be in sine.
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!
Last edited by weaknuclearforce; 23 Feb 2017 at 4:50 PM.
I think I thought that at first but the square root has plus not minus so I just did trig substitution
Thanks, didn't realise this.
Thanks for the help anyway, but just a note to integrate secx just multiply top and bottom by (secx+tanx)
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Buy my books/notes cheaply here!
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Uni Course: Actuarial Studies and Statistics at MQ -- PM me if you have questions
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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.)
Last edited by InteGrand; 23 Feb 2017 at 6:02 PM.
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