Solving an Anti-Derivative Problem with Trigonometric Identities

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find the anti-derivative of :

f(x) = 4 - 3(1+x^2)^(-1)

I have thought this question for hours...but no clue at all...

that is what I have attempted:

F(x) = 4x - 3 Ln(1 + x^2) ...
but if i differentiated it ---
then I got F'(x) = 4 - 3*(2x)/(1 + x^2)...

is there anyway to eliminate the (2x) ...?

please help
 
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Well, first the "4" is obvious- you are completely correct that it's anti-derivative is 4x.

Now, for 3/(1+ x^2). Log doesn't work because 1+ x^2 is not x!

Do you know the derivative of arctan(x)?
 
oh...thank you...
I totally forgot about there are some formula for that...
yes...
so is the answer 4x - 3 arc tan x + C?
 
but in my textbook, i can not find the prove of those identities...
just wonder if there is any link for that...
 
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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