Expert Tips for Integrating sqrt(1+4x^2) | Simplify with u=2x+1

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To integrate sqrt(1+4x^2), a user suggests rewriting it as sqrt(-4x+(2x + 1)^2) and substituting u=2x+1, but finds this approach unhelpful. Another participant recommends using partial integration directly with sqrt(1+4x^2)dx and applying the (+1 -1) trick in the integrand's numerator. An alternative substitution of u=2x can simplify the process, though it's not mandatory. The discussion highlights the importance of using hyperbolic identities and trigonometric substitutions to facilitate the integration. Overall, the conversation emphasizes various strategies to tackle the integral effectively.
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I am trying to integrate sqrt(1+4*x^2)
I have been trying to rewrite this into sqrt(-4x+(2x + 1)^2) and putting u=2x+1 and substituting but I don't think that makes this any easier. Could someone please give me a hint.
 
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Swatch said:
I am trying to integrate sqrt(1+4*x^2)
I have been trying to rewrite this into sqrt(-4x+(2x + 1)^2) and putting u=2x+1 and substituting but I don't think that makes this any easier. Could someone please give me a hint.


The answer is Partial Integration...Give it a try and let me know

marlon

edit hint : do the partial integration right a way with the sqrt(1+4x^2)dx. After this you will need to apply the (+1 -1) trick in the integrand's numerator. If you want you can first do the substitution u=2x to get rid of the 4, but it is not compulsory...
 
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After some hard work and a lot of eraser I got the right answer. Thank you marlon.
 
Also:

\int\sqrt{1+\left(2x\right)^2}\,dx=\frac{1}{2}\int\cosh^2{x}\,dx

Then use the identity for the double angle to simplify that integral.
 
Try x=\frac{1}{2}\tan \theta \Rightarrow \sqrt{1+4x^2}=\sec\theta.
 
Question: A clock's minute hand has length 4 and its hour hand has length 3. What is the distance between the tips at the moment when it is increasing most rapidly?(Putnam Exam Question) Answer: Making assumption that both the hands moves at constant angular velocities, the answer is ## \sqrt{7} .## But don't you think this assumption is somewhat doubtful and wrong?

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