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

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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.
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...

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