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Question Number 43419 by Raj Singh last updated on 10/Sep/18

Commented by maxmathsup by imad last updated on 10/Sep/18
$let A = ∫ ((xdx)/((x−1)(x^2 +1)^2 )) A = ∫ ((x−1+1)/((x−1)(x^2 +1)^2 ))dx = ∫ (dx/(x^2 +1)) + ∫ (dx/((x−1)(x^2 +1)^2 )) but ∫ (dx/(x^2 +1)) =arctanx +c_1 changemene x −1=t give ∫ (dx/((x−1)(x^2 +1)^2 )) = ∫ (dt/(t{ (t+1)^2 +1}^2 )) =_(t+1 =tanθ) ∫ (((1+tan^2 θ)dθ)/((tanθ−1){1+tan^2 θ}^2 )) = ∫ ((cos^2 θ)/(((sinθ)/(cosθ))−1)) dθ = ∫ ((cos^3 θ)/(sinθ −cosθ)) dθ = ∫ ((cos^3 θ)/((√2)cos(θ +(π/4)))) dθ =_(θ +(π/4)=u) ∫ ((cos^3 (u−(π/4)))/((√2)cosu)) du = ∫ (({((cosu)/(√2))+((sinu)/(√2))}^3 )/((√2)cosu)) du = (1/4) ∫ (((cosu +sinu)^3 )/(cosu)) du =(1/4) ∫ ((cos^3 u +3cos^2 u sinu +3cosu sin^2 u +sin^3 u)/(cosu)) du =(1/4) ∫ cos^2 u du +(3/4) ∫ cosu sinu du +(3/4) ∫ sin^2 u du +(1/4) ∫ ((sin^3 u)/(cosu)) du =(1/8) ∫ (1+cos(2u)du +(3/8) ∫ sin(2u)du +(3/8) ∫ (1−cos(2u))du +(1/4) ∫((sin^3 u)/(cosu))du = (1/8) u +(1/(16))sin(2u) −(3/(16))cos(2u) +(3/8)u −(3/(16))sin(2u) +(1/4) ∫ (((1−cos^2 u)sinu)/(cosu))du u=θ +(π/4) =arctan(t+1) +(π/4) ....be contiued...$
$l e t A = \int \frac{x d x}{(x - 1) {(x^{2} + 1)}^{2}}$ $A = \int \frac{x - 1 + 1}{(x - 1) {(x^{2} + 1)}^{2}} d x = \int \frac{d x}{x^{2} + 1} + \int \frac{d x}{(x - 1) {(x^{2} + 1)}^{2}} b u t$ $\int \frac{d x}{x^{2} + 1} = a r c t a n x + c_{1} c h a n g e m e n e x - 1 = t g i v e$ $\int \frac{d x}{(x - 1) {(x^{2} + 1)}^{2}} = \int \frac{d t}{t {{(t + 1)}^{2} + 1}^{2}} =_{t + 1 = t a n θ} \int \frac{(1 + {t a n}^{2} θ) d θ}{(t a n θ - 1) {1 + {t a n}^{2} θ}^{2}}$ $= \int \frac{{c o s}^{2} θ}{\frac{s i n θ}{c o s θ} - 1} d θ = \int \frac{{c o s}^{3} θ}{s i n θ - c o s θ} d θ = \int \frac{{c o s}^{3} θ}{\sqrt{2} c o s (θ + \frac{π}{4})} d θ$ $=_{θ + \frac{π}{4} = u} \int \frac{{c o s}^{3} (u - \frac{π}{4})}{\sqrt{2} c o s u} d u = \int \frac{{\frac{c o s u}{\sqrt{2}} + \frac{s i n u}{\sqrt{2}}}^{3}}{\sqrt{2} c o s u} d u$ $= \frac{1}{4} \int \frac{{(c o s u + s i n u)}^{3}}{c o s u} d u$ $= \frac{1}{4} \int \frac{{c o s}^{3} u + 3 {c o s}^{2} u s i n u + 3 c o s u {s i n}^{2} u + {s i n}^{3} u}{c o s u} d u$ $= \frac{1}{4} \int {c o s}^{2} u d u + \frac{3}{4} \int c o s u s i n u d u + \frac{3}{4} \int {s i n}^{2} u d u + \frac{1}{4} \int \frac{{s i n}^{3} u}{c o s u} d u$ $= \frac{1}{8} \int (1 + c o s (2 u) d u + \frac{3}{8} \int s i n (2 u) d u + \frac{3}{8} \int (1 - c o s (2 u)) d u + \frac{1}{4} \int \frac{{s i n}^{3} u}{c o s u} d u$ $= \frac{1}{8} u + \frac{1}{16} s i n (2 u) - \frac{3}{16} c o s (2 u) + \frac{3}{8} u - \frac{3}{16} s i n (2 u) + \frac{1}{4} \int \frac{(1 - {c o s}^{2} u) s i n u}{c o s u} d u$ $u = θ + \frac{π}{4} = a r c t a n (t + 1) + \frac{π}{4} . . . . b e c o n t i u e d . . .$
Commented by maxmathsup by imad last updated on 10/Sep/18
$another way for the integral ∫ (dx/((x−1)(x^2 +1)^2 )) we decompose the fraction F(x) = (1/((x−1)(x^2 +1)^2 )) F(x) =(a/(x−1)) +((bx+c)/(x^2 +1)) +((dx +e)/((x^2 +1)^2 )) a=lim_(x→1) (x−1)F(x) =(1/2) lim_(x→+∞) x F(x) =0 =a+b ⇒b=−(1/2) ⇒ F(x) = (1/(2(x−1))) −(1/2) ((x−2c)/(x^2 +1)) + ((dx+e)/((x^2 +1)^2 )) we have F(0) =−1 =−(1/2) +c +e ⇒c+e =−(1/2) F(2)= (1/2) −(1/2) ((2−2c)/5) + ((2d +e)/(25)) =(1/(25)) F(−1) = −(1/8) =−(1/4) −(1/2) ((−1−2c)/2) +((−d +e)/4) weget a linear system with unknown c ,d,e ....$
$a n o t h e r w a y f o r t h e i n t e g r a l \int \frac{d x}{(x - 1) {(x^{2} + 1)}^{2}} w e d e c o m p o s e t h e$ $f r a c t i o n F (x) = \frac{1}{(x - 1) {(x^{2} + 1)}^{2}}$ $F (x) = \frac{a}{x - 1} + \frac{b x + c}{x^{2} + 1} + \frac{d x + e}{{(x^{2} + 1)}^{2}}$ $a = {l i m}_{x \to 1} (x - 1) F (x) = \frac{1}{2}$ ${l i m}_{x \to + \infty} x F (x) = 0 = a + b \Rightarrow b = - \frac{1}{2} \Rightarrow$ $F (x) = \frac{1}{2 (x - 1)} - \frac{1}{2} \frac{x - 2 c}{x^{2} + 1} + \frac{d x + e}{{(x^{2} + 1)}^{2}} w e h a v e$ $F (0) = - 1 = - \frac{1}{2} + c + e \Rightarrow c + e = - \frac{1}{2}$ $F (2) = \frac{1}{2} - \frac{1}{2} \frac{2 - 2 c}{5} + \frac{2 d + e}{25} = \frac{1}{25}$ $F (- 1) = - \frac{1}{8} = - \frac{1}{4} - \frac{1}{2} \frac{- 1 - 2 c}{2} + \frac{- d + e}{4} w e g e t a l i n e a r s y s t e m w i t h u n k n o w n$ $c, d, e . . . .$
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