Finding the Anti-Derivative of (2+x^2)/(1+x^2): A Scientific Approach

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Homework Statement



f'(x)= (2+x^2)/(1+x^2) Find anti derivative

Homework Equations


The Attempt at a Solution



I attempted to bring the denominator up using (1+x^2)^-1 and i also tried long division to simplify but had no luck...

1/(1+x^2) is the inverse tan derivative, but what can i do from here:

(2+x^2) * 1/(1+x^2) is substitution legal here?
 
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Try putting (2+x^2)/(1+x^2) as 2/(1+x^2) + x^2/(1+x^2)

then divide out the second fraction
 
Or write
\frac{2+x^2}{1+ x^2}= \frac{1}{1+x^2}+ \frac{1+ x^2}{1+ x^2}
 
so i would cancel the second term then take the anti derivate to be left with invesre tan of x + x + C .. is this correct?
 
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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