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Maths integral

HI guys, i could not find an answer to this integral, so i was just wondering if someone could be so kind and check through my working and see if i did it right

Many thanks in advance.
8BB7DCDC-5BF7-4C37-90CD-9BA92DEE260A.jpg178.5KB
Original post by yash Kainth
HI guys, i could not find an answer to this integral, so i was just wondering if someone could be so kind and check through my working and see if i did it right

Many thanks in advance.

Having had a quick look: after your subsitution you have to evaluate the integral
∫1u2−4du\displaystyle\int \frac{1}{\sqrt{u^2-4}}du
and you seem to (incorrectly) state that this is equal to
u2−4u\frac{\sqrt{u^2-4}}{u}.

FYI, you can check answers for integrals using Wolfram Alpha. Just type "integrate x/sqrt(x^2-2x+5)". Try not to abuse this (i.e. make sure you actually do your work rather than using Wolfram Alpha to cheat!).
Reply 2
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YGSK
OP
13
Thank you so much!
Reply 3
Avatar for davros
davros
16
Original post by yash Kainth
HI guys, i could not find an answer to this integral, so i was just wondering if someone could be so kind and check through my working and see if i did it right

Many thanks in advance.

I would start by completing the square inside the square root in the denominator. The substitution u = x - 1 will then transform the integrand into something of the form

u+1u2+a2\dfrac{u + 1}{\sqrt{u^2 + a^2}}

where a is a constant you can determine when you complete the square.

You can then split the integral into the sum of 2 pieces - the first is easily evaluated with a further substitution, and the second can be recognised as a standard integral :smile:
Reply 4
Avatar for simon0
simon0
13
HINT: Note the derivative of x2−2x+5, x ^{2} -2x + 5, (with respect to x) is 2(x−1), 2(x-1), so try having the numerator as:

(x−1)+1. \displaystyle (x-1) + 1.

As @davros says, then split up the integral where to one substitution may be an option and to the other use standard results (completing the square and looking at the formula booklet may be helpful here).

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