Math Question and Answers


Question and Answers Forum

All Questions Topic List
Arithmetic Questions
Previous in All Question Next in All Question
Previous in Arithmetic Next in Arithmetic
Question Number 167776 by peter frank last updated on 24/Mar/22

	Answered by MikeH last updated on 24/Mar/22

	${a x}^{2} + b x + c = a (x^{2} + \frac{b}{a} x + \frac{c}{a})$ $= a [{(x + \frac{b}{2 a})}^{2} - \frac{b^{2}}{4 a^{2}} + \frac{c}{a}]$ $= a [{(x + \frac{b}{2 a})}^{2} + \frac{- b^{2} + 4 a c}{4 a^{2}}]$ $= a {(x + \frac{b}{2 a})}^{2} + \frac{4 a c - b^{2}}{4 a}$ $\int \frac{1}{a {(x + \frac{b}{2 a})}^{2} + \frac{4 a c - b^{2}}{4 a}} d x$ $Let u = x + \frac{b}{2 a} \Rightarrow d u = d x$ $\int \frac{1}{{a u}^{2} + k} d u, where k = \frac{4 a c - b^{2}}{4 a}$ $\int \frac{1}{a (u^{2} + {(\sqrt{\frac{k}{a}})}^{2})} d u = \frac{1}{a} \sqrt{\frac{a}{k}} \tan^{- 1} (x \sqrt{\frac{a}{k}}) + p .$
	Commented by peter frank last updated on 26/Mar/22

	$thank you$
	Answered by MJS_new last updated on 25/Mar/22

	$\int \frac{d x}{{a x}^{2} + b x + c} =$ $[t = 2 a x + b \to d x = \frac{d t}{2 a}]$ $= 2 \int \frac{d t}{t^{2} + 4 a c - b^{2}}$ $(1) 4 a c - b^{2} < 0 \Rightarrow 4 a c - b^{2} = - d^{2} \land d > 0$ $2 \int \frac{d t}{t^{2} + 4 a c - b^{2}} = 2 \int \frac{d t}{t^{2} - d^{2}} = 2 \int \frac{d t}{(t - d) (t + d)} =$ $= \frac{1}{d} \int (\frac{1}{t - d} - \frac{1}{t + d}) d t = \frac{\ln (t - d) - \ln (t + d)}{d} =$ $= \frac{\ln ∣ 2 a x + b - \sqrt{b^{2} - 4 a c} ∣ - \ln ∣ 2 a x + b + \sqrt{b^{2} - 4 a x} ∣}{\sqrt{b^{2} - 4 a c}} + C$ $(2) 4 a c - b^{2} = 0$ $2 \int \frac{d t}{t^{2}} = - \frac{2}{t} = - \frac{2}{2 a x + b} + C$ $(3) 4 a c - b^{2} > 0 \Rightarrow 4 a c - b^{2} = d^{2} \land d > 0$ $2 \int \frac{d t}{t^{2} + 4 a c - b^{2}} = 2 \int \frac{d t}{t^{2} + d^{2}} =$ $= \frac{2 \arctan \frac{t}{d}}{d} =$ $= \frac{2 \arctan \frac{2 a x + b}{\sqrt{4 a c - b^{2}}}}{\sqrt{4 a c - b^{2}}} + C$
	Commented by peter frank last updated on 26/Mar/22

	$thank you$
	Commented by MJS_new last updated on 27/Mar/22

	$hey Peter, {you}^{'} re welcome! regards from$ $Vienna (Austria), Martin$
Terms of Service
Privacy Policy
Contact: info@tinkutara.com

Question and Answers Forum