Differentiate the following functions with respect to x :

$\frac{1}{a x^{2} + b x + c}$

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Solution

Let $y = \frac{1}{a x^{2} + b x + c}$

Differentiating y w.r.t. x, we get

$\frac{d y}{d x} = \frac{(a x^{2} + b x + c) \frac{d}{d x} (1) - 1 \times \frac{d}{d x} (a x^{2} + b x + c)}{(a x^{2} + b x + c)^{2}}$ (By quotient formula)

$= \frac{(a x^{2} + b x + c) \times 0 - 1 \times (2 a x + b)}{(a x^{2} + b x + c)^{2}}$

$\Rightarrow \frac{d y}{d x} = - \frac{(2 a x + b)}{(a x^{2} + b x + c)^{2}}$

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\int R (x, \sqrt{a x^{2} + b x + c}) d x

I. \sqrt{a x^{2} + b x + c} = t \pm x \sqrt{a}

a > 0;

II. \sqrt{a x^{2} + b x + c} = t x \pm \sqrt{c}

c > 0;

III. \sqrt{a x^{2} + b x + c} = (x - a) t \pm x

a x^{2} + b x + c = a (x - α) (x - β)

α

a x^{2} + b x + c = 0.

\frac{d x}{x + \sqrt{x^{2} - x + 1}}

t

x ?

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