( x x+1 + x-1 d x=A(x)(x+1)^ 3 2 +B(x)(x-1)^ 3 2 +C ) , then (A(x)+B(x)= )

Step-by-step solution

( aligned & Let I= x x+1 + x-1 d x & = x[ x+1 - x-1 ] x+1-x+1 d x & = 1 2 x x+1 d x- 1 2 x x-1 d x= 1 2 I₁- 1 2 I₂ aligned ) Now, (I₁= x x+1 d x ) Put (x+1=u d x=d u ) ( aligned I₁ & = (u-1) u d x= (u^ 3 2 -u^ 1 2 ) d x & = 2 5 u^ 5 2 - 2 3 u^ 3 2 +c₁= 2 5 (x+1)^ 5 2 - 2 3 (x+1)^ 3 2 +c₁ & =2(x+1)^ 3 2 [ 1 5 (x+1)- 1 3 ]+c₁ & =2(x+1)^ 3 2 [ 3 x+3-5 15 ]+c₁ & = 2(3 x-2) 15 (x+1)^ 3 2 +c₁ aligned ) Again, (I₂= x x-1 d x ) Put (x-1=v ) ( aligned & d x=d v & I₂= (v+1) v d v= (v^ 3 2 +v^ 1 2 ) d v = & 2 5 v^ 5 2 + 2 3 v