Question

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

Options

• `1/6`

• `1/30`

• `-1/30`

• `-1/6`

Answer 1

If `int 1/((x^2 + 4)(x^2 + 9))dx = A tan^-1  x/2 + B tan^-1(x/3) + C`, then A – B = `underlinebb(1/6)`.

Explanation:

Consider I = `int 1/((x^2 + 4)(x^2 + 9))dx`

= `int1/5(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[1/2 tan^-1  x/2 - 1/3 tan^-1  x/3] + C`

= `1/10  tan^-1  x/2 - 1/15 tan^-1  x/3 + C`

We have, I = `A tan^-1  x/2 + Btan^-1  x/3 + C`

So, `Atan^-1  x/2 + B tan^-1  x/3 + C`

= `1/10 tan^-1  x/2 - 1/15 tan^-1  x/3 + C`

After comparing on both sides, we get

A = `1/10` and B = `(-1)/15`

Hence, A – B = `1/10 + 1/15 = (15 + 10)/150 = 1/6`