If `int (dx)/((x+2)(x^(2)+1))=a"In"(1+x^(2))+btan^(-1)x+(1)/(5)"In"|x+2|+C.` Then

Options

A. `a= (-1)/(10), b = (-2)/(5)`
B. `a = (1)/(10) , b = -2/5`
C. `a = (-1)/(10),b = (2)/(5)`
D. `a = 1/10 , b = 2/5`

Answer

A

Explanation

Given that, `int(dx)/((x+2)(x^(2)+1)) = a log|1+x^(2)|+b tan^(-1) x + 1/5 log|x+2| + C`
Now, ` I = int (dx)/((x+2)(x^(2) +1))`
`rArr 1 = A (x^(2) +1) + (Bx + C) (x+2)`
`rArr 1 = Ax^(2)+A + Bx^(2)+2Bx+Cx+2C`
`rArr 1= (A + B)x^(2) + (2 B + C)x + A + 2C`
` rArr A + B = 0, A + 2C = 1, 2B + C = 0`
We have, `A = 1/5 , B = - 1/5` and `C = 2/5`
`:. int (dx)/((x+2)(x^(2) +1)) = 1/5 int (1)/(x+2)dx + int (-(1)/(5)x+2/5)/(x^(2)+1)dx`
`=1/5int(1)/(x+2)dx-1/5int(x)/(1+x^(2))dx+1/5int(2)/(1+x^(2))dx`
`= 1/5 log|x+2|-1/10log|1+x^(2)|+2/5tan^(-1)x+C`
`:. b = 2/5` and `a = (-1)/(10)`