Integrating 1/sqrt(x^2+16): Common Mistakes and Correct Solution

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Homework Statement



Integrate 1/sqrt(x^2+16)


The answer I get is :
ln ((sqrt(16+x^2)+x)/4)+C

and it is incorrect. The answer is ln(sqrt(16+x^2)+x) + C
 
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Hint:

ln(a/b)=ln(a)-ln(b)
 
but ln4 is not =0
 
No, but C- ln(4) is still just a constant. Basically, your solution is correct but it is also equivalent to the other solution just with a different value of C.
 
Of course! Thanks a lot:)
 

Thread 'Prove that the integral is equal to ##\pi^2/8##'

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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