Math Question and Answers


Question and Answers Forum

All Questions Topic List
Algebra Questions
Previous in All Question Next in All Question
Previous in Algebra Next in Algebra
Question Number 156170 by MathSh last updated on 08/Oct/21

	Answered by mindispower last updated on 09/Oct/21

	$H e l l o s i r w i t h e r e s o d u e [t h e o r m i s d i r a c t l y$ $\frac{p (x)}{q (x)} = d e g q > d e g (p) + 1 i n t h i s c a s e$
	Commented by MathSh last updated on 09/Oct/21

	$Hola dear Ser, thank you$ $how, if possible please$
	Answered by MJS_new last updated on 09/Oct/21
	$∫(dx/((x^2 +x+1)(a^2 x^2 +1)))= =(1/(a^4 −a^2 +1))∫((a^2 x+1)/(x^2 +x+1))dx−(a^2 /(a^4 −a^2 +1))∫((a^2 x−a^2 +1)/(a^2 x^2 +1))dx (1/(a^4 −a^2 +1))∫((a^2 x+1)/(x^2 +x+1))dx= =(a^2 /(2(a^4 −a^2 +1)))∫((2x+1)/(x^2 +x+1))dx−((a^2 −2)/(2(a^4 −a^2 +1))∫(dx/(x^2 +x+1))= =(a^2 /(2(a^4 −a^2 +1)))ln (x^2 +x+1) −(((√3)(a^2 −2))/(3(a^4 −a^2 +1)))arctan (((√3)(2x+1))/3) −(a^2 /(a^4 −a^2 +1))∫((a^2 x−a^2 +1)/(a^2 x^2 +1))dx= =−(a^4 /(a^4 −a^2 +1))∫(x/(a^2 x^2 +1))dx+((a^2 (a^2 −1))/(a^4 −a^2 +1))∫(dx/(a^2 x^2 +1))= =−(a^2 /(2(a^4 −a^2 +1)))ln (a^2 x^2 +1) +((a(a^2 −1))/(a^4 −a^2 +1))arctan (ax) inserting the borders I get Ω(a)= { (((1/(a^4 −a^2 +1))((π/(18))(9a^3 −2(√3)a^2 −9a+4(√3))−a^2 ln a)∧a>0)),((((2(√3))/9)π∧a=0)),((−(1/(a^4 −a^2 +1))((π/(18))(9a^3 +2(√3)a^2 −9a−4(√3))+a^2 ln (−a))∧a<0)) :}$
	$\int \frac{d x}{(x^{2} + x + 1) (a^{2} x^{2} + 1)} =$ $= \frac{1}{a^{4} - a^{2} + 1} \int \frac{a^{2} x + 1}{x^{2} + x + 1} d x - \frac{a^{2}}{a^{4} - a^{2} + 1} \int \frac{a^{2} x - a^{2} + 1}{a^{2} x^{2} + 1} d x$ $\frac{1}{a^{4} - a^{2} + 1} \int \frac{a^{2} x + 1}{x^{2} + x + 1} d x =$ $= \frac{a^{2}}{2 (a^{4} - a^{2} + 1)} \int \frac{2 x + 1}{x^{2} + x + 1} d x - \frac{a^{2} - 2}{2 (a^{4} - a^{2} + 1} \int \frac{d x}{x^{2} + x + 1} =$ $= \frac{a^{2}}{2 (a^{4} - a^{2} + 1)} \ln (x^{2} + x + 1) - \frac{\sqrt{3} (a^{2} - 2)}{3 (a^{4} - a^{2} + 1)} \arctan \frac{\sqrt{3} (2 x + 1)}{3}$ $- \frac{a^{2}}{a^{4} - a^{2} + 1} \int \frac{a^{2} x - a^{2} + 1}{a^{2} x^{2} + 1} d x =$ $= - \frac{a^{4}}{a^{4} - a^{2} + 1} \int \frac{x}{a^{2} x^{2} + 1} d x + \frac{a^{2} (a^{2} - 1)}{a^{4} - a^{2} + 1} \int \frac{d x}{a^{2} x^{2} + 1} =$ $= - \frac{a^{2}}{2 (a^{4} - a^{2} + 1)} \ln (a^{2} x^{2} + 1) + \frac{a (a^{2} - 1)}{a^{4} - a^{2} + 1} \arctan (a x)$ $inserting the borders I get$ $Ω (a) = {\begin{cases} \frac{1}{a^{4} - a^{2} + 1} (\frac{π}{18} (9 a^{3} - 2 \sqrt{3} a^{2} - 9 a + 4 \sqrt{3}) - a^{2} \ln a) \land a > 0 \\ \frac{2 \sqrt{3}}{9} π \land a = 0 \\ - \frac{1}{a^{4} - a^{2} + 1} (\frac{π}{18} (9 a^{3} + 2 \sqrt{3} a^{2} - 9 a - 4 \sqrt{3}) + a^{2} \ln (- a)) \land a < 0 \end{cases}$
	Commented by MathSh last updated on 09/Oct/21

	$Perfect my dear S er, thank you$
Terms of Service
Privacy Policy
Contact: info@tinkutara.com

Question and Answers Forum