If `int(dx)/((x+2)(x^(2)+1)) = alog|1+x^(2)|+btan^(-1)x+ 1/5log|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

C

Explanation

`int (dx)/((x+2)(x^(2)+1))=a"In"(1+x^(2))+btan^(-1)x+(1)/(5)"In"|x+2|+C`
Differentiating both sides, we get
` (1)/((x+2)(x^(2)+1))=(2ax)/((1+x^(2)))+(b)/((1+x^(2)))+(1)/(5(x+2))`
or `(1)/((x+2)(x^(2)+1))=((x+2)(5b+10ax)+1+x^(2))/(5(1+x^(2))(x+2))`
or `5=(1+x^(2))+5(b+2ax)(x+2)`
Comparing the like powers of x on both sides, we get
`1+10a=0, b+4a=0, 10b+1=5`
or `a= -(1)/(10), b=(2)/(5)`