Question

Resolver:
∫ (5x+2)/(x²+7x+10) dx

Answer 1

∫ (5x+2)/(x²+7x+10) dx

Descompondremos (5x+2)/(x²+7x+10) en sus fracciones parciales:

Factoricemos el denominador:
x²+7x+10 = (x+2)(x+5)

(5x+2)/(x²+7x+10) = A/(x+2) + B/(x+5) = (A(x+5) + B(x+2))/(x²+7x+10)

 Fijémonos en los numeradores:

A(x+5) + B(x+2) = 5x+2
Ax + 5A + Bx + 2B = 5x+2
x(A+B) + (5a + 2B) = 5x+2

A+B = 5..........B= 5-A
5A + 2B = 2

Resolviendo:
5A + 2(5-A) = 2
5A + 10 - 2A = 2
3A  = -8
A= -8/3


B= 5- (-8/3)
B= 23/3

Retomamos:
(5x+2)/(x²+7x+10)= 23/(3(x+5))- 8/(3(x+2))

∫ (5x+2)/(x²+7x+10) dx= ∫  23/(3(x+5))- 8/(3(x+2)) dx

∫  23/(3(x+5))- 8/(3(x+2)) dx = ∫  23/(3(x+5)) dx - ∫ 8/(3(x+2)) dx
∫ 23/(3(x+5)) dx = 23/3 ∫ 1/(x+5) dx
∫ 1/(x+5) dx, u= x+5
du/dx= 1, du=dx
∫ 1/(u) du = Ln (u)=Ln(x+5)

∫ 8/(3(x+2)) dx = 8/3∫ 1/(x+2) dx
∫ 1/(x+2) dx, u= x+2
du/dx = 1, du= dx
∫ 1/(u) du= ln (u)= Ln (x+2)

Entonces:

∫ (5x+2)/(x²+7x+10) dx = 23/3(Ln(x+5)) - 8/3(Ln (x+2)) + C