The value of ₀^ x (1+x) (x²+1 ) d x is

Step-by-step solution

Let I= ₀^ x d x (1+x) (x²+1 ) By partial fraction, x (1+x) (x²+1 ) = A (1+x) + B x+C (x²+1 ) x=A (x²+1 )+(1+x)(B x+C) x=A (x²+1 )+ (B x+B x²+C+C x ) x=(A+B) x²+(B+C) x+(A+C) On comparing both sides, we get A+B=0, B+C=1, A+C=0 ( i ) On adding all these equations, we get aligned A+B+C &= 1 2 ...(ii) A= 1 2 -1 &=- 1 2 , C= 1 2 and B= 1 2 I= ₀^ -1 2(1+x) + 1 2 (x+1) (x²+1 ) d x =- 1 2 ₀^ d x 1+x + 1 2 ₀^ x x²+1 d x + 1 2 ₀^ d x 1+x² aligned =- 1 2 [ (1+x)]₀^ + 1 4 [ (x²+1 ) ]₀^ + 1 2 2 =- 1 2 _ x (1+x)+ 1 4 _ x (1+x² )