calculate ∫_0 ^1 (((x^2 −1)ln(x))/((x^2 +2x−1)(x^2 −2x−1)))dx


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Question Number 47114 by maxmathsup by imad last updated on 04/Nov/18

$c a l c u l a t e \int_{0}^{1} \frac{(x^{2} - 1) l n (x)}{(x^{2} + 2 x - 1) (x^{2} - 2 x - 1)} d x$
Commented by maxmathsup by imad last updated on 11/Nov/18
$let decompose F(x)=((x^2 −1)/((x^2 +2x−1)(x^2 −2x−1))) ⇒F(x)=((x^2 −1)/((x^2 +2x+1−2)(x^2 −2x+1 −2))) =((x^2 −1)/(((x+1)^2 −2)((x−1)^2 −2))) =((x^2 −1)/((x+1−(√2))(x+1+(√2))(x−1−(√2))(x−1+(√2)))) =((x^2 −1)/((x−x_1 )(x−x_2 )(x−t_1 )(x−t_2 ))) with x_1 =−1+(√2), x_2 =−1−(√2), t_1 =1+(√2) and t_2 =1−(√2) F(x)=(a/(x−x_1 )) +(b/(x−x_2 )) +(c/(x−t_1 )) +(d/(x−t_2 )) a =lim_(x→x_(1 ) ) (x−x_1 )F(x)=((x_(1 ) ^2 −1)/((x_1 −x_2 )(x_1 −t_1 )(x_1 −t_2 ))) =((2−2(√2))/(2(√2)(−2)(−2+2(√2)))) =((−1)/(−4(√2))) =(1/(4(√2))) . b =lim_(x→x_2 ) (x−x_2 )F(x) = ((x_(2 ) ^2 −1)/((x_2 −x_1 )(x_2 −t_1 )(x_2 −t_2 ))) =((2+2(√2))/(−2(√2))(−2−2(√2))(−2))) =−(1/(4(√2))) c =lim_(x→t_1 ) (x−t_1 )F(x) =((t_1 ^2 −1)/((t_1 −x_1 )(t_1 −x_2 )(t_1 −t_2 ))) =((2+2(√2))/(2(2+2(√2))2(√2))) =(1/(4(√2))) d =lim_(x→t_2 ) (x−t_2 )F(t) = ((t_2 ^2 −1)/((t_2 −x_1 )(t_2 −x_2 )(t_2 −t_1 ))) =((2−2(√2))/(2−2(√2))(2)(−2(√2)))) =−(1/(4(√2))) ⇒ F(x) = (1/(4(√2))){ (1/(x−x_1 )) −(1/(x−x_2 )) +(1/(x−t_1 )) −(1/(x−t_2 ))} ⇒ I =∫_0 ^1 ln(x)F(x)dx =(1/(4(√2))) {∫_0 ^1 ((ln(x)dx)/(x−x_1 )) −∫_0 ^1 ((ln(x))/(x−x_2 ))dx +∫_0 ^1 ((ln(x))/(x−t_1 ))dx−∫_0 ^1 ((ln(x))/(x−t_2 ))dx} let determine ∫_0 ^1 ((ln(x))/(x−x_1 ))dx....be continued....$
$l e t d e c o m p o s e F (x) = \frac{x^{2} - 1}{(x^{2} + 2 x - 1) (x^{2} - 2 x - 1)}$ $\Rightarrow F (x) = \frac{x^{2} - 1}{(x^{2} + 2 x + 1 - 2) (x^{2} - 2 x + 1 - 2)} = \frac{x^{2} - 1}{({(x + 1)}^{2} - 2) ({(x - 1)}^{2} - 2)}$ $= \frac{x^{2} - 1}{(x + 1 - \sqrt{2}) (x + 1 + \sqrt{2}) (x - 1 - \sqrt{2}) (x - 1 + \sqrt{2})} = \frac{x^{2} - 1}{(x - x_{1}) (x - x_{2}) (x - t_{1}) (x - t_{2})}$ $w i t h x_{1} = - 1 + \sqrt{2}, x_{2} = - 1 - \sqrt{2}, t_{1} = 1 + \sqrt{2} a n d t_{2} = 1 - \sqrt{2}$ $F (x) = \frac{a}{x - x_{1}} + \frac{b}{x - x_{2}} + \frac{c}{x - t_{1}} + \frac{d}{x - t_{2}}$ $a = {l i m}_{x \to x_{1}} (x - x_{1}) F (x) = \frac{x_{1}^{2} - 1}{(x_{1} - x_{2}) (x_{1} - t_{1}) (x_{1} - t_{2})}$ $= \frac{2 - 2 \sqrt{2}}{2 \sqrt{2} (- 2) (- 2 + 2 \sqrt{2})} = \frac{- 1}{- 4 \sqrt{2}} = \frac{1}{4 \sqrt{2}} .$ $b = {l i m}_{x \to x_{2}} (x - x_{2}) F (x) = \frac{x_{2}^{2} - 1}{(x_{2} - x_{1}) (x_{2} - t_{1}) (x_{2} - t_{2})}$ $= \frac{2 + 2 \sqrt{2}}{- 2 \sqrt{2}) (- 2 - 2 \sqrt{2}) (- 2)} = - \frac{1}{4 \sqrt{2}}$ $c = {l i m}_{x \to t_{1}} (x - t_{1}) F (x) = \frac{t_{1}^{2} - 1}{(t_{1} - x_{1}) (t_{1} - x_{2}) (t_{1} - t_{2})}$ $= \frac{2 + 2 \sqrt{2}}{2 (2 + 2 \sqrt{2}) 2 \sqrt{2}} = \frac{1}{4 \sqrt{2}}$ $d = {l i m}_{x \to t_{2}} (x - t_{2}) F (t) = \frac{t_{2}^{2} - 1}{(t_{2} - x_{1}) (t_{2} - x_{2}) (t_{2} - t_{1})}$ $= \frac{2 - 2 \sqrt{2}}{2 - 2 \sqrt{2}) (2) (- 2 \sqrt{2})} = - \frac{1}{4 \sqrt{2}} \Rightarrow$ $F (x) = \frac{1}{4 \sqrt{2}} {\frac{1}{x - x_{1}} - \frac{1}{x - x_{2}} + \frac{1}{x - t_{1}} - \frac{1}{x - t_{2}}} \Rightarrow$ $I = \int_{0}^{1} l n (x) F (x) d x = \frac{1}{4 \sqrt{2}} {\int_{0}^{1} \frac{l n (x) d x}{x - x_{1}} - \int_{0}^{1} \frac{l n (x)}{x - x_{2}} d x + \int_{0}^{1} \frac{l n (x)}{x - t_{1}} d x - \int_{0}^{1} \frac{l n (x)}{x - t_{2}} d x}$ $l e t d e t e r m i n e \int_{0}^{1} \frac{l n (x)}{x - x_{1}} d x . . . . b e c o n t i n u e d . . . .$
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