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Resolver :∫ (5x + 2) / (x² + 7x + 10) dx?

Resolver : ∫ (5x + 2) / (x² + 7x + 10) dx.

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A1103

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∫ (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.