Question

Can anyone help me, please? `intdx/(x^2-4x-13`

Answer 1

`int \ 1/(x^2-4x-13) \ dx = 1/(2sqrt(17)) \ { ln|x-2-sqrt(17)| -ln|x-2+sqrt(17)| } + C`

Explanation:

We have:

`I = int \ 1/(x^2-4x-13) \ dx`

We can decompose the integrand into partial fraction. First note that the roots of the equation (using the quadratic formula) are:

`x^2-4x-13 =0 => x=2+-sqrt(17)`

So we can factorise the denominator, and decompose into partial fractions, which will take the form:

`1/(x^2-4x-13) -= 1/( (x-2-sqrt(17))(x-2+sqrt(17)) )`
`" " = A/(x-2-sqrt(17)) + B/(x-2+sqrt(17))`
`" " = (A(x-2+sqrt(17)) + B(x-2-sqrt(17))) / ((x-2-sqrt(17))(x-2+sqrt(17)))`

Leading to the identity:

`1 -= A(x-2+sqrt(17)) + B(x-2-sqrt(17))`

Where `A,B` are constants that are to be determined. We can find them by substitutions (In practice we do this via the "cover up" method:

Put `x = 2+sqrt(17) => 1 = \ \ \ \ \ 2Asqrt(17) => A = \ \ \ \ \ 1/(2sqrt(17))`
Put `x = 2-sqrt(17) => 1 = -2Bsqrt(17) => B = -1/(2sqrt(17))`

So using partial fraction decomposition we have:

`I = int \ (1/(2sqrt(17)))/(x-2-sqrt(17)) + (-1/(2sqrt(17)))/(x-2+sqrt(17)) \ dx`
`\ \ = 1/(2sqrt(17)) \ int \ 1/(x-2-sqrt(17)) -1/(x-2+sqrt(17)) \ dx`

And now all integrands are readily integratable, so:

`I = 1/(2sqrt(17)) \ { ln|x-2-sqrt(17)| -ln|x-2+sqrt(17)| } + C`

Answer 2

If you're willing to use inverse hyperbolic trigonometric functions, see below.

Explanation:

`int 1/(x^2-4x-13) dx = int 1/((x-2)^2-17) dx`

`= 1/17 int 1/(((x-2)/sqrt17)^2-1) dx`

Let `u = (x-2)/sqrt17` so `du = 1/sqrt17 dx` and the integral becomes:

`1/sqrt17 int 1/(u^2-1) du = -1/sqrt17 int 1/(1-u^2) du`

`= -1/sqrt17 tanh^-1 u +C`

`= -1/sqrt17 tanh^-1((x-2)/sqrt17) +C`