Question

`int dx/((x+2)(x^2 + 1))`    ...(given)

`1/(x^2 +1) dx = tan ^-1 + c`

Answer 1

`int 1/((x+2)(x^2 + 1))`

`=A/(x+2) + (Bx + C)/((x^2+1))`

`therefore 1 = A(x^2 + 1 ) + (Bx + C)(x+2)`

Put x= -2 , we get `A=1/5`

Now comparing the coefficients of x² and constant term, we get

`0 =A+B`

`0 = 1/5 + B`

`therefore B = -1/5`

and `1 = A + 2C`

`therefore 1 = 1/5 + 2C`

`therefore 2C = 4/5` 

`therefore C = 2/5`

`therefore B = -1/5, C= 2/5`

`1/((x+2)(x^2 + 1)) = (1/5)/((x+2)) + (1/5x + 2/5)/((x^2 + 1))`

`I = 1/5 int dx/(x+2)-1/10 int (2x)/(x^2 + 1)dx + 2/5 intdx/(x^2 +1)`

`I = 1/5 log |x+2|- 1/10 log |x^2 + 1|+ 2/5 tan^-1(x) + c`