Ex 7.5, 7

Transcript

Ex 7.5, 7 𝑥/(𝑥^2 + 1)(𝑥 − 1) Let I=∫1▒𝑥/(𝑥^2 + 1)(𝑥 − 1) 𝑑𝑥 We can write integrand as 𝑥/(𝑥^2 + 1)(𝑥 − 1) =(𝐴𝑥 + 𝐵)/((𝑥^2 + 1) ) + 𝐶/((𝑥 − 1) ) 𝑥/(𝑥^2 + 1)(𝑥 − 1) =((𝐴𝑥 + 𝐵)(𝑥 − 1) + 𝐶(𝑥^2 + 1))/(𝑥^2 + 1)(𝑥 − 1) By Cancelling denominator 𝑥=(𝐴𝑥+𝐵)(𝑥−1)+𝐶(𝑥^2+1) Putting x = 1, in (1) 𝑥=(𝐴𝑥+𝐵)(𝑥−1)+𝐶(𝑥^2+1) 1=(𝐴×1+𝐵)(1−1)+𝐶(1^2+1) 1=(𝐴+𝐵)×0+𝐶(1+1) 1=0+2𝐶 𝐶=1/2 Putting x = 0 , in (1) 𝑥=(𝐴𝑥+𝐵)(𝑥−1)+𝐶(𝑥^2+1) 0=(𝐴×0+𝐵)(0−1)+𝐶(0+1) 0 =𝐵(−1)+𝐶 ⇒𝐵=𝐶=1/2 Putting x = −1 −1=(𝐴(−1)+𝐵)(−1−1)+𝐶((1)^2+1) −1=(−𝐴+𝐵)(−2)+𝐶(1+1) −1=2𝐴−2𝐵+2𝐶 −1=2𝐴−2×1/2+2×1/2 −1=2𝐴−1+1 −1=2𝐴 𝐴=(−1)/2 Hence we can write 𝑥/(𝑥^2 + 1)(𝑥 − 1) =((− 1/2 𝑥 + 1/2))/((𝑥^2 + 1) ) + (1/2)/((𝑥 − 1) ) =(−1 . 𝑥)/(2 (𝑥^2 + 1) ) + 1/(2 (𝑥^2 + 1) )+1/2(𝑥 − 1) Integrating 𝑤.𝑟.𝑡.𝑥 I=∫1▒𝑥/(𝑥^2 + 1)(𝑥 − 1) 𝑑𝑥 =−1/2 ∫1▒𝑥/(𝑥^2 + 1) 𝑑𝑥+1/2 ∫1▒𝑑𝑥/(𝑥^2 + 1) + 1/2 ∫1▒𝑑𝑥/(𝑥 − 1) 𝑑𝑥 Solving 𝐈𝟏 I1=−1/2 ∫1▒𝑥/((𝑥^2 + 1) ) 𝑑𝑥 Put 𝑥^2+1=𝑡 Different both sides w.r.t.𝑥 2𝑥+0=𝑑𝑡/𝑑𝑥 𝑑𝑥=𝑑𝑡/2𝑥 Hence we can write −1/2 ∫1▒𝑥/((𝑥^2 + 1) ) 𝑑𝑥=−1/2 ∫1▒𝑥/𝑡 𝑑𝑡/2𝑥 =−1/(2 × 2) ∫1▒𝑑𝑡/𝑡 =−1/4 log⁡|𝑡|+𝐶1 =−1/4 log⁡|𝑥^2+1|+𝐶1 Solving 𝐈𝟐 I2=1/2 ∫1▒1/(𝑥^2 + 1) 𝑑𝑥 =1/2 tan^(−1)⁡𝑥+𝐶2 Solving 𝐈𝟑 I3=1/2 ∫1▒1/(𝑥 − 1) 𝑑𝑥 =1/2 log⁡|𝑥−1|+𝐶3 Hence I=I1+I2+I3 =−1/4 log⁡|𝑥^2+1|+𝐶1+1/2 tan^(−1)⁡𝑥+𝐶2+1/2 log⁡|𝑥−1|+𝐶3 =−1/4 log⁡|𝑥^2+1|+1/2 tan^(−1)⁡𝑥+1/2 log⁡|𝑥−1|+𝐶 =−1/4 log⁡|𝑥^2+1|+1/2 tan^(−1)⁡𝑥+1/2 log⁡|𝑥−1|+𝐶 =𝟏/𝟐 𝒍𝒐𝒈⁡|𝒙−𝟏|−𝟏/𝟒 〖𝒍𝒐𝒈 〗⁡|𝒙^𝟐+𝟏|+𝟏/𝟐 〖𝒕𝒂𝒏〗^(−𝟏)⁡𝒙+ C