There exist positive integers A , B , C such that ∫ 1 2 x 2 − 1 4 x 2 + 1 x d x = A 1 π + ln B − tan − 1 C . What is the value of A + B + C ?
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Nice. Comparing this with my chain of substitutions, I have w = 4 x 2 + 1 x 2 − 1 , so we are pretty much on the same page.
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@Patrick Corn , I read your answer. It is useful and self-explanatory. But when i put this integrand in my hp 50g calculator, i have to turn on 'approx mode' and 'complex mode' to get the answer 1 . 9 3 1 1 8 8 4 6 6 8 3 ( π / 2 − arctan ( 3 ) + ln ( 5 ) ) . In 'Real mode' It answers 'Infinite results'.
Does that mean the answer to this question is imaginary?
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No, it just means that your calculator is not perfect at evaluating improper integrals. In approximate more, it is probably trying to evaluate the integral from 1 + ε to X where ε > 0 is very small and X is very large. It may be that it uses a different numerical algorithm in Real mode to Complex mode, which works better in this case.
it is easy to find integrals that (a) exist, (b) cannot be handled by handheld calculators. It is still not too difficult to beat Mathematica in the same way.
The successive substitutions u 2 = 4 x 2 + 1 , u = 5 sec v , 2 w = sin v give ∫ x 2 − 1 4 x 2 + 1 x d x = 2 ∫ ( u 2 − 1 ) u 2 − 5 u 2 d u = 2 ∫ u 2 − 5 d u + 2 ∫ ( u 2 − 1 ) u 2 − 5 d u = 2 cosh − 1 ( 5 u ) + 2 ∫ 4 + sin 2 v cos v d v = 2 cosh − 1 ( 5 4 x 2 + 1 ) + ∫ 1 + w 2 d w = 2 cosh − 1 ( 5 4 x 2 + 1 ) + tan − 1 w + k = 2 cosh − 1 ( 5 4 x 2 + 1 ) + tan − 1 ( 4 x 2 + 1 x 2 − 1 ) + k for an arbitrary constant k , and hence ∫ 1 2 x 2 − 1 4 x 2 + 1 x d x = [ 2 cosh − 1 ( 5 4 x 2 + 1 ) + tan − 1 ( 4 x 2 + 1 x 2 − 1 ) ] 1 2 = 2 cosh − 1 5 3 + tan − 1 3 1 = 2 1 π + ln 5 − tan − 1 3
making the answer 2 + 5 + 3 = 1 0 .
@Mark Hennings , Let F ( x ) = x 1 x 2 − 1 4 x 2 + 1 , Then, F ( 1 ) = ∞ . Now, what is the use of computing integral of this function F ( x ) ?
What is the logic behind selecting u = 5 sec v and 2 w = sin v ? Your answer in red color is difficult to understand. Why don't you edit it using latex formatting so that reader, viewers can understand it quickly and in easiest way.
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There are plenty of examples of finite integrals of functions that tend to ∞ at a point. Consider ∫ 0 1 x 1 d x for example. These are what are known as improper integrals.
This integral, or an extension of it, was useful in solving another problem I posted at the same time - hence the title.
As to the substitutions, I was successively making the integral a little easier each time, until I ended up with something I could handle.
As to the red answer, occasionally (or not so occasionally) the Brilliant system develops glitches, so that what is well-written LaTeX is not parsed correctly, and needs to be retyped. Take that problem up with the Brilliant people, not me.
Substituting x = cosh ( t ) .
∫ a b sinh ( t ) 4 cosh 2 ( t ) + 1 cosh ( t ) sinh ( t ) d t = ∫ a b cosh ( t ) 4 cosh 2 ( t ) + 1 d t where a = 0 and b = cosh − 1 ( 2 ) . Wolfram alpha helped me on this, resulting in
2 sinh − 1 ( ( 5 ) 2 sinh ( t ) ) + tan − 1 ( 2 cosh ( 2 t ) + 3 sinh ( t ) ) which evaluates to 0 for the lower bound a = 0 , so that we just need the value at b.
Since b is positive, sinh ( b ) = cosh 2 ( b ) − 1 = 2 − 1 = 1
Also, cosh ( 2 t ) = 2 cosh 2 ( t ) − 1 , so our answer simplifies into
2 sinh − 1 ( 5 2 ) + tan − 1 ( 4 cosh 2 ( b ) + 1 1 )
Filling in b = cosh − 1 ( ( 2 ) ) shows that 4 cosh 2 ( b ) + 1 = 3 , and
Solving sinh ( y ) = 5 2 ⟹ ( e y ) 2 − 5 4 e y − 1 = 0 ⟹ e y = 5 ⟹ y = ln ( 5 ) (discarding the imaginary solution) ⟹ 2 sinh − 1 ( 5 2 ) = 2 ln ( 5 ) = ln ( 5 )
Finally t a n − 1 ( x 1 ) = 2 π − t a n − 1 ( x ) so we arrive at
l n ( 5 ) + 2 π − tan − 1 ( 3 ) The requested answer is 2 + 5 + 3 = 1 0 .
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Let u = x 2 − 1 4 x 2 + 1 . Then u 2 − 4 = x 2 − 1 5 , so x 2 = u 2 − 4 u 2 + 1 and 2 x d x = ( u 2 − 4 ) 2 − 1 0 u d u .
Hence x d x = ( x d x ) x 2 1 = ( u 2 − 4 ) 2 − 5 u d u u 2 + 1 u 2 − 4 = − ( u 2 + 1 ) ( u 2 − 4 ) 5 u d u .
Then the integral becomes ∫ 3 ∞ ( u 2 + 1 ) ( u 2 − 4 ) 5 u 2 d u = ∫ 3 ∞ ( u 2 + 1 1 + u 2 − 4 4 ) d u = ( arctan ( u ) + ln ( u + 2 u − 2 ) ) ∣ ∣ ∣ ∣ 3 ∞ = π / 2 − ( arctan ( 3 ) + ln ( 3 + 2 3 − 2 ) ) = π / 2 − arctan ( 3 ) − ln ( 1 / 5 ) = 2 π − arctan ( 3 ) + ln ( 5 ) . So the answer is 2 + 5 + 3 = 1 0 .