Menu Close

Question-43322

Tinku Tara 0 Comments
Pin




Question Number 43322 by Raj Singh last updated on 09/Sep/18
Commented by maxmathsup by imad last updated on 11/Sep/18
let A = ∫ (dx/((x+1)^2 (x^2 +1))) let decompose F(x) = (1/((x+1)^2 (x^2 +1))) F(x) = (a/(x+1)) +(b/((x+1)^2 )) +((cx +d)/(x^2 +1)) b =lim_(x→−1) (x+1)^2 F(x) =(1/2) lim_(x→+∞) xF(x) =0 =a+c ⇒c =−(1/2) ⇒F(x) =(a/(x+1)) +(1/(2(x+1)^2 )) +((−(1/2)x +d)/(x^2 +1)) F(o) =1 = a +(1/2) +d ⇒ a+d =(1/2) F(2) =(1/(45)) = (a/3) +(1/(18)) +((d−1)/5) ⇒ 1 =15a +((45)/(18)) +9d−9 ⇒ 1 =15a +(5/2) +9d−9 ⇒15a +9d =10−(5/2) =((15)/2) ⇒ 15a +9((1/2)−a)=((15)/2) ⇒6a =((15)/2) −(9/2) =3 ⇒a=(1/2) ⇒d=0 ⇒ F(x)= (1/(2(x+1))) +(1/(2(x+1)^2 )) −(x/(2(1+x^2 ))) ⇒ A = ∫ F(x)dx =(1/2)ln∣x+1∣ −(1/(2(x+1))) −(1/4)ln(1+x^2 ) +c .
letA=∫dx(x+1)2(x2+1)letdecomposeF(x)=1(x+1)2(x2+1)F(x)=ax+1+b(x+1)2+cx+dx2+1b=limx→−1(x+1)2F(x)=12limx→+∞xF(x)=0=a+c⇒c=−12⇒F(x)=ax+1+12(x+1)2+−12x+dx2+1F(o)=1=a+12+d⇒a+d=12F(2)=145=a3+118+d−15⇒1=15a+4518+9d−9⇒1=15a+52+9d−9⇒15a+9d=10−52=152⇒15a+9(12−a)=152⇒6a=152−92=3⇒a=12⇒d=0⇒F(x)=12(x+1)+12(x+1)2−x2(1+x2)⇒A=∫F(x)dx=12ln∣x+1∣−12(x+1)−14ln(1+x2)+c.
Answered by tanmay.chaudhury50@gmail.com last updated on 09/Sep/18
2)∫(dx/((x+1)^2 (x^2 +1))) ∫(a/((x+1)))dx+∫(b/((x+1)^2 ))dx+∫((cx+d)/(x^2 +1))dx now calculating a b c and d (1/((x+1)^2 (x^2 +1)))=(a/(x+1))+(b/((x+1)^2 ))+((cx+d)/(x^2 +1)) 1=a(x+1)(x^2 +1)+b(x^2 +1)+(cx+d)(x+1)^2 1=a(x^3 +x^2 +x+1)+b(x^2 +1)+(cx+d)(x^2 +2x+1) 1=a(x^3 +x^2 +x+1)+b(x^2 +1)+(cx^3 +2cx^2 +cx+dx^2 +2dx+d) 1=x^3 (a+c)+x^2 (a+b+2c+d)+x(a+c+2d)+(a+b+d) a+c==0 a+b+2c+d=0 a+c+2d=0 a+b+d=1 a+c+d=0 0+d=0 d=0 a+b=1 a+b+2c+d=0 1+2c+0=0 c=((−1)/2) a+c=0 a=−c a=(1/2) a=(1/2) a+b=1 b=(1/2) c=((−1)/2) d=0 ∫(a/(x+1))dx+∫(b/((x+1)^2 ))+∫((cx)/(x^2 +1))dx (1/2)∫(dx/(x+1))+(1/2)∫(dx/((x+1)^2 )) + ((−1)/2)∫(dx/(x^2 +1)) (1/2)ln(x+1)+(1/2)×(((x+1)^(−1) )/(−1))−(1/2)tan^(−1) (x)+c (1/2)ln(x+1)−(1/(2(x+1)))−(1/2)tan^(−1) (x)+c
2)∫dx(x+1)2(x2+1)∫a(x+1)dx+∫b(x+1)2dx+∫cx+dx2+1dxnowcalculatingabcandd1(x+1)2(x2+1)=ax+1+b(x+1)2+cx+dx2+11=a(x+1)(x2+1)+b(x2+1)+(cx+d)(x+1)21=a(x3+x2+x+1)+b(x2+1)+(cx+d)(x2+2x+1)1=a(x3+x2+x+1)+b(x2+1)+(cx3+2cx2+cx+dx2+2dx+d)1=x3(a+c)+x2(a+b+2c+d)+x(a+c+2d)+(a+b+d)a+c==0a+b+2c+d=0a+c+2d=0a+b+d=1a+c+d=00+d=0d=0a+b=1a+b+2c+d=01+2c+0=0c=−12a+c=0a=−ca=12a=12a+b=1b=12c=−12d=0∫ax+1dx+∫b(x+1)2+∫cxx2+1dx12∫dxx+1+12∫dx(x+1)2+−12∫dxx2+112ln(x+1)+12×(x+1)−1−1−12tan−1(x)+c12ln(x+1)−12(x+1)−12tan−1(x)+c

Pin

Leave a Reply

Your email address will not be published. Required fields are marked *