If `int(1)/((x^(2)+4)(x^(2)+9))dx=A" tan"^(-1)(x)/(2)+Btan^(-1)((x)/(3))+C`, then A-B=

Options

A. `1/6`
B. `1/30`
C. `-1/30`
D. `-1/6`

Answer

A

Explanation

Let `I=int(1)/((x^(2)+4)(x^(2)+9))dx`
`= int(1)/((x^(2) +-x^(2)-4))[(1)/(x^(2) -4)-(1)/(x^(2) + 9)]dx`
`=int((1)/(x^(2)+4)-(1)/(x^(2)+9))dx`
`=1/5[int(1)/(x^(2)+2^(2))dx-int(1)/(x^(2)+3^(2))dx]`
`=1/5[1/2tan^(-1)'x/2-1/3tan^(-1)'x/3]+C`
`= 1/10 tan^(-1)'x/2-1/15 tan^(-1)'x/3+C`
But, it is given that, `I = A tan^(-1)'x/2 + B tan^(-1)'x/3 + C`
`A tan^(-1) 'x/2 +Btan^(-1)'x/3+C`
`= 1/10 tan^(-1) 'x/2-1/15 tan^(-1)'x/3+C`
`= 1/10 tan^(-1)x/2- 1/15 tan^(-1)'x/3 + C`
On comparing both sides , we get
`A= 1/(10)` and `B = (-1)/(15)`
Now, `A - B = 1/10 + 1/15`
`rArr (15 + 10 )/(150) = (25)/(150) = 1/6`