∫((7x−6)/((x^2 +25)(√((x−3)^2 +4)))) dx = ?


Question and Answers Forum

All Questions Topic List
Integration Questions
Previous in All Question Next in All Question
Previous in Integration Next in Integration
Question Number 35909 by ajfour last updated on 25/May/18

$\int \frac{7 x - 6}{(x^{2} + 25) \sqrt{{(x - 3)}^{2} + 4}} d x = ?$
Commented by tanmay.chaudhury50@gmail.com last updated on 26/May/18

$t h i s n u t i s h a r d t o c r a c k . . . i a m f i g h t i n g . . h o p e$ $r e a c h t h e d e s t i n a t i o n t o g e t i t s o l v e d . . .$
Commented by rahul 19 last updated on 26/May/18

$Y e s s i r!$ $I f t h e r e i s n o (7 x - 6) i n n u m e r a t o r$ $t h e n i t i s c a k e w a l k!$
Commented by prof Abdo imad last updated on 26/May/18

$t b e r e i s n o c a k e n i c h o c o l a t i n m a t h s . . .$
Commented by abdo mathsup 649 cc last updated on 27/May/18
$changement x−3=2sht give I = ∫ ((7(3+2sht)−6)/({ (2sht+3)^2 +25}2 cht)) 2chtdt I = ∫ ((14sht +15)/({4sh^2 t +12sht +34)))dt = ∫ ((14((e^t −e^(−t) )/2) +15)/(4 (((e^t −e^(−t) )/2))^2 +12 ((e^t −e^(−t) )/2) +34))dt = ∫ ((7e^t −7 e^(−t) +15)/(e^(2t) +e^(−2t) −2 +6 e^t −6 e^(−t) +34))dt =∫ ((7e^t −7 e^(−t) +15)/(e^(2t) +e^(−2t) +6 e^t −6 e^(−t) +32))dt changement e^t =x give I = ∫ ((7x−(7/x) +15)/(x^2 +(1/x^2 ) +6x −(6/x) +32)) (dx/x) = ∫ ((7x^2 +15x −7)/(x^4 +1 +6x^3 −6x +32x^2 ))dx = ∫ ((7x^2 +15x −7)/(x^4 +6x^3 +32x^2 −6x +1))dx the roots of polynom p(x)= x^4 +6x^3 +32 x^2 −6x +1 are the complex z_1 ∼ −3,093+4,853 i z_2 ∼−3,093 −4,853 i =z_1 ^− z_3 ∼ 0,093 +0,146 i z_4 ∼0,093 −0,146 i= z_3 ^− p(x) ∼ (z−_ z_1 )(x−z_1 ^− )(x−z_2 )(x −z_2 ^− ) ∼ (x^2 +2.3,093x+(√((3,093)^2 +(4,853)^2 ))). ( x^2 +2.0,093 x +(√((0,093)^2 +(0,146)^2 ))) the value of this integral is obtained after decomposing the fraction F(x) = ((7x^2 +15x −7)/(p(x))) at form F(x) = ((ax+b)/(x^2 −2Re(z_1 )x +∣z_1 ∣^2 )) + ((cx+d)/(x^2 −2Re(z_2 )x +∣z_2 ∣^( )) ....$
$c h a n g e m e n t x - 3 = 2 s h t g i v e$ $I = \int \frac{7 (3 + 2 s h t) - 6}{{{(2 s h t + 3)}^{2} + 25} 2 c h t} 2 c h t d t$ $I = \int \frac{14 s h t + 15}{{4 {s h}^{2} t + 12 s h t + 34)} d t$ $= \int \frac{14 \frac{e^{t} - e^{- t}}{2} + 15}{4 {(\frac{e^{t} - e^{- t}}{2})}^{2} + 12 \frac{e^{t} - e^{- t}}{2} + 34} d t$ $= \int \frac{7 e^{t} - 7 e^{- t} + 15}{e^{2 t} + e^{- 2 t} - 2 + 6 e^{t} - 6 e^{- t} + 34} d t$ $= \int \frac{7 e^{t} - 7 e^{- t} + 15}{e^{2 t} + e^{- 2 t} + 6 e^{t} - 6 e^{- t} + 32} d t c h a n g e m e n t$ $e^{t} = x g i v e$ $I = \int \frac{7 x - \frac{7}{x} + 15}{x^{2} + \frac{1}{x^{2}} + 6 x - \frac{6}{x} + 32} \frac{d x}{x}$ $= \int \frac{7 x^{2} + 15 x - 7}{x^{4} + 1 + 6 x^{3} - 6 x + 32 x^{2}} d x$ $= \int \frac{7 x^{2} + 15 x - 7}{x^{4} + 6 x^{3} + 32 x^{2} - 6 x + 1} d x t h e r o o t s o f$ $p o l y n o m p (x) = x^{4} + 6 x^{3} + 32 x^{2} - 6 x + 1 a r e$ $t h e c o m p l e x z_{1} \sim - 3, 093 + 4, 853 i$ $z_{2} \sim - 3, 093 - 4, 853 i = {\bar{z}}_{1}$ $z_{3} \sim 0, 093 + 0, 146 i$ $z_{4} \sim 0, 093 - 0, 146 i = {\bar{z}}_{3}$ $p (x) \sim (z -_{} z_{1}) (x - {\bar{z}}_{1}) (x - z_{2}) (x - {\bar{z}}_{2})$ $\sim (x^{2} + 2.3, 093 x + \sqrt{{(3, 093)}^{2} + {(4, 853)}^{2}}) .$ $(x^{2} + 2.0, 093 x + \sqrt{{(0, 093)}^{2} + {(0, 146)}^{2}})$ $t h e v a l u e o f t h i s i n t e g r a l i s o b t a i n e d a f t e r$ $d e c o m p o s i n g t h e f r a c t i o n$ $F (x) = \frac{7 x^{2} + 15 x - 7}{p (x)} a t f o r m$ $F (x) = \frac{a x + b}{x^{2} - 2 R e (z_{1}) x + ∣ z_{1} ∣^{2}} + \frac{c x + d}{x^{2} - 2 R e (z_{2}) x + ∣ z_{2} ∣^{(}}$ $. . . .$
	Answered by tanmay.chaudhury50@gmail.com last updated on 26/May/18
	$∫((7x−6)/((x^2 +25)(√(x^2 −6x+13))))dx let t^2 =((x^2 −6x+13)/(x^2 +25)) 2lnt=ln(x^2 −6x+13)−ln(x^2 +25) (2/t)dt={((2x−6)/((x^2 −6x+13)))−((2x)/(x^2 +25))}dx (2/t)dt=((2x^3 +50x−6x^2 −150−2x^3 +12x^2 −26x)/((x^2 +25)(x^2 −6x+13)))dx (2/t)dt=((6x^2 +24x−150)/((x^2 +25)(x^2 −6x+13)))dx dx=(2/t)×(((x^2 +25)(x^2 −6x+13))/(6(x^2 +4x−25)))dt ∫((7x−6)/((x^2 +25)(√(x^2 −6x+13))))(2/t)×(((x^2 +25)(x^2 −6x+13)/(6(x^2 +4x−25))) contd puting value of t ∫((7x−6)/((x^2 +25)(√(x^2 −6x+13))))×((2(√(x^2 +25)))/(√(x^2 −6x+13)))× (((x^2 +25)(x^2 −6x+13))/(6(x^2 +4x−25)))dx (1/3)∫(((7x−6)(√(x^2 +25)))/(x^2 +4x−25))dx contd$
	$\int \frac{7 x - 6}{(x^{2} + 25) \sqrt{x^{2} - 6 x + 13}} d x$ $l e t t^{2} = \frac{x^{2} - 6 x + 13}{x^{2} + 25}$ $2 l n t = l n (x^{2} - 6 x + 13) - l n (x^{2} + 25)$ $\frac{2}{t} d t = {\frac{2 x - 6}{(x^{2} - 6 x + 13)} - \frac{2 x}{x^{2} + 25}} d x$ $\frac{2}{t} d t = \frac{2 x^{3} + 50 x - 6 x^{2} - 150 - 2 x^{3} + 12 x^{2} - 26 x}{(x^{2} + 25) (x^{2} - 6 x + 13)} d x$ $\frac{2}{t} d t = \frac{6 x^{2} + 24 x - 150}{(x^{2} + 25) (x^{2} - 6 x + 13)} d x$ $d x = \frac{2}{t} \times \frac{(x^{2} + 25) (x^{2} - 6 x + 13)}{6 (x^{2} + 4 x - 25)} d t$ $\int \frac{7 x - 6}{(x^{2} + 25) \sqrt{x^{2} - 6 x + 13}} \frac{2}{t} \times \frac{(x^{2} + 25) (x^{2} - 6 x + 13}{6 (x^{2} + 4 x - 25)}$ $c o n t d p u t i n g v a l u e o f t$ $\int \frac{7 x - 6}{(x^{2} + 25) \sqrt{x^{2} - 6 x + 13}} \times \frac{2 \sqrt{x^{2} + 25}}{\sqrt{x^{2} - 6 x + 13}} \times$ $\frac{(x^{2} + 25) (x^{2} - 6 x + 13)}{6 (x^{2} + 4 x - 25)} d x$ $\frac{1}{3} \int \frac{(7 x - 6) \sqrt{x^{2} + 25}}{x^{2} + 4 x - 25} d x$ $c o n t d$
	Commented by ajfour last updated on 26/May/18

	$t h a n k s f o r t h e a t t e m p t s i r .$
Terms of Service
Privacy Policy
Contact: info@tinkutara.com

Question and Answers Forum