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Question Number 110970 by khaki last updated on 01/Sep/20

Commented by khaki last updated on 01/Sep/20

$please solve this queztion$
	Answered by mr W last updated on 01/Sep/20

	$\int \frac{a x + b}{{a x}^{2} + b x + c} d x$ $= \frac{1}{2} \int \frac{2 a x + b + b}{{a x}^{2} + b x + c} d x$ $= \frac{1}{2} [\int \frac{2 a x + b}{{a x}^{2} + b x + c} d x + \int \frac{b}{{a x}^{2} + b x + c} d x]$ $= \frac{1}{2} \int \frac{d ({a x}^{2} + b x + c)}{{a x}^{2} + b x + c} + \frac{1}{2} \int \frac{b}{{a x}^{2} + b x + c} d x$ $= \frac{1}{2} \ln ∣ {a x}^{2} + b x + c ∣ + \frac{b}{2} \int \frac{1}{{a x}^{2} + b x + c} d x$ $= . . . .$
	Commented by mr W last updated on 01/Sep/20

	Commented by Her_Majesty last updated on 01/Sep/20

	$w h a t i f b^{2} = 4 a c ?$
	Commented by malwan last updated on 01/Sep/20

	$i f b^{2} = 4 a c \Rightarrow Δ = 0$ $\Rightarrow x = \frac{- b}{2 a}$ $\int \frac{d x}{{a x}^{2} + b x + c} = \int \frac{d x}{{(x + \frac{b}{2 a})}^{2}}$ $= \int {(x + \frac{b}{2 a})}^{- 2} d (x + \frac{b}{2 a}) = \frac{- 1}{x + \frac{b}{2 a}}$
	Answered by malwan last updated on 01/Sep/20

	$\int \frac{a x + b}{{a x}^{2} + b x + c} d x$ $= \int \frac{a x + \frac{1}{2} b}{{a x}^{2} + b x + c} d x - \frac{b}{2} \int \frac{d x}{{a x}^{2} + b x + c}$ $= \frac{1}{2} l n ∣ {a x}^{2} + b x + c ∣ - \frac{b}{2} I$ $I = \int \frac{d x}{a [x^{2} + \frac{b}{a} x + \frac{c}{a}]}$ $= \frac{1}{a} \int \frac{d x}{[{(x + \frac{b}{2 a})}^{2} + (\frac{c}{a} - \frac{b^{2}}{4 a^{2}})]}$ $= \frac{1}{a} \int \frac{d x}{[{(\frac{2 a x + b}{2 a})}^{2} + \frac{4 a c - b^{2}}{4 a^{2}}]}$ $= \frac{1}{a} \frac{1}{\frac{4 a c - b^{2}}{4 a^{2}}} \int \frac{d x}{\frac{{(2 a x + b)}^{2}}{4 a^{2}} / \frac{4 a x - b^{2}}{4 a^{2}} + 1}$ $= \frac{4 a}{4 a c - b^{2}} \int \frac{d x}{{(\frac{2 a c + b}{\sqrt{4 a c - b^{2}}})}^{2} + 1}$ $t = \frac{2 a c}{\sqrt{4 a c - b^{2}}} \Rightarrow d t \sqrt{4 a c - b^{2}} = 2 a d x$ $d x = \frac{\sqrt{4 a c - b^{2}}}{2 a}$ $∴ I = \frac{4 a}{4 a c - b^{2}} \int \frac{\frac{\sqrt{4 a c - b^{2}}}{2 a} d t}{t^{2} + 1}$ $= \frac{2}{\sqrt{4 a c - b^{2}}} {t a n}^{- 1} t + C$ $I = \frac{2}{\sqrt{4 a c - b^{2}}} {t a n}^{- 1} (\frac{2 a x + b}{\sqrt{4 a c - b^{2}}}) + C$
	Commented by mr W last updated on 01/Sep/20

	$i t h i n k$ $= \frac{1}{2} l n ∣ {a x}^{2} + b x + c ∣ + \frac{b}{2} I$
	Commented by malwan last updated on 01/Sep/20

	$y e s S i r$
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