Question

Evaluate the following integrals:

`int (2x + 1)/(x^2 + 4x - 5).dx`

Answer 1

Let I = `int (2x + 1)/(x^2 + 4x - 5).dx`

Let 2x + 1 = `"A"[d/dx(x^2 + 4x - 5)] + "B"`

2x + 1 = A(2x + 4) + B

∴ 2x + 1 = 2Ax + (4A + B)

Comparing the coefficient of x and constant on both sides, we get,

2A = 2	and	4A + B = 1
∴ A = 1	and	∴  4(1) + B = 1
		∴ B = 1 - 4
		∴ B = - 3

∴ 2x + 1 = (2x + 1) - 3

∴ I = `int ((2x + 1) - 3)/(x^2 + 4x + 5)."dx"`

∴ I = `int (2x + 1)/(x^2 + 4x - 5)."dx" - 3 int (1)/(x^2 + 4x - 5)."dx"`

∴ I = `"I"_1 - 3"I"_2`

I₁ is of the type `int (f'(x))/f(x).dx = log|f(x)| + c`

∴ `"I"_1 = log|x^2 + 4x - 5| + c_1`

∴  I₂ = `int (1)/(x^2 + 4x - 5).dx`

∴  I₂ = `int (1)/((x^2 + 4x + 4) - 4 - 5).dx`

∴  I₂ = `int (1)/((x + 2)^2 - 3^2).dx`

∴  I₂ = `1/(2 × 3) log |(x + 2 - 3)/(x + 2 + 3)| + c_2`

∴  I₂ = `1/6 log |(x - 1)/(x + 5)| + c_2`

∴ I = `log|x^2 + 4x - 5| - 3 × 1/6 log|(x - 1)/(x + 5)| + c`.

∴ I = `log|x^2 + 4x - 5| - 1/2 log|(x - 1)/(x + 5)| + c`.