Question

`int "dx"/(x^2 + 4x + 13)` = ?

Options

• `1/3 tan^-1 ((x + 2)/3)` + C

• `1/6 log ((x - 1)/(x + 5))` + C

• `3 tan^-1 ((x + 2)/3)` + C

• `1/6 tan^-1 ((x + 2)/3)` + C

Answer 1

`1/3 tan^-1 ((x + 2)/3)` + C

Explanation:

We have,

I = `int "dx"/(x^2 + 4x + 13)`

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

I = `int "dx"/((x + 2)^2 + (3)^2)`

I = `1/3 tan^-1 ((x + 2)/3)` + C