Question

`int_0^1 1/((x^2 + 16)(x^2 + 25))dx` = ______.

Options

• `1/5[1/4 tan^-1(1/4) - 1/5 tan^-1(1/5)]`

• `1/9[1/4 tan^-1(1/4) - 1/5 tan^-1(1/5)]`

• `1/4[1/4 tan^-1(1/4) - 1/5 tan^-1(1/5)]`

• `1/9[1/5 tan^-1(1/4) - 1/5 tan^-1(1/5)]`

Answer 1

`int_0^1 1/((x^2 + 16)(x^2 + 25))dx` = `underlinebb(1/9[1/4 tan^-1(1/4) - 1/5 tan^-1(1/5)])`.

Explanation:

Let I = `int_0^1 (dx)/((x^2 + 16)(x^2 + 25))`

= `1/9int_0^1(1/(x^2 + 16) - 1/(x^2 + 25))dx`

= `1/9(1/4tan^-1  x/4 - 1/5 tan^-1  x/5)_0^1`

= `1/9[1/4tan^-1  1/4 - 1/5 tan^-1  1/5]`