How Do You Derive the Integral of 1/(x^2 + a^2)?

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Kalie
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Yeah this one is good...

Derive the following:

int(1/(x^2 + a^2)dx = 1/a*arctan(x/a) + C

any idea how I would derive this thing?
Cause I'm totally lost...
 
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Well do you agree that if you show that the derivative of 1/a*arctan(x/a) + C is 1/(x^2 + a^2, then you will have attained your goal?
 
fundamental theorem of calculus
 
no he actually wants us to plow thorugh the intregal using subsitution and stuff and then somehow end up at the answer
 
use the substitution x = a\tan \theta
 

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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