Verified answer

Hola

ʃ (x - 3) dx/(2 x^2 + 6 x) = (1/2) ʃ (x - 3) dx/(x^2 + 3 x)

El divisor se puede considerar como

x^2 + 3 x = x^2 + 2 (3/2) x

x^2 + 3 x = x^2 + 2 (3/2) x + (3/2)^2 - (3/2)^2

x^2 + 3 x = (x + (3/2))^2 - (3/2)^2

Ahora sustituimos

u = x + (3/2)

x = u - (3/2)

x - 3 = u - (3/2) - 3 = u - (9/2)

du = dx

(1/2) ʃ (x - 3) dx/(x^2 + 3 x) = (1/2) ʃ (u - (9/2)) du/(u^2 - (3/2)^2)

(1/2) ʃ (x - 3) dx/(x^2 + 3 x) = (1/2) ʃ (u - (3/2) - 3) du/(u^2 - (3/2)^2)

(1/2) ʃ (x - 3) dx/(x^2 + 3 x) = (1/2) ʃ (u - (3/2)) du/(u^2 - (3/2)^2) -

- (3/2) ʃ du/(u^2 - (3/2)^2)

diferencia de cuadrados

u^2 - (3/2)^2 = (u - (3/2))(u + (3/2))

La primera integral queda

(1/2) ʃ (u - (3/2)) du/(u^2 - (3/2)^2) = (1/2) ʃ du/(u + (3/2))

La segunda integral

implica tratar la fracción

1/( (u - (3/2))(u + (3/2)) )

recordamos que

(u + (3/2)) - (u - (3/2)) = 3

(1/3) (u + (3/2)) - (u - (3/2)) = 1

Entonces

1/( (u - (3/2))(u + (3/2)) ) =

= (1/3) (u + (3/2)) - (u - (3/2))/( (u - (3/2))(u + (3/2)) )

= (1/3) (1/(u - (3/2)) - (1/3) (1/(u + (3/2))

La segunda integral queda

- (3/2) ʃ du/(u^2 - (3/2)^2)

= - (1/2) ʃ du/(u - (3/2)) + (1/2) ʃ du/(u + (3/2))

En total

(1/2) ʃ (x - 3) dx/(x^2 + 3 x) = (1/2) ʃ (u - (3/2)) du/(u^2 - (3/2)^2) -

- (3/2) ʃ du/(u^2 - (3/2)^2)

= (1/2) ln(u + (3/2) - (1/2) ln(u - (3/2)) + (1/2) ln(u + (3/2)) + C

= ln(u + (3/2)) - (1/2) ln(u - (3/2)) + C

= ln(x + (3/2) + (3/2)) - (1/2) ln(x + (3/2) - (3/2)) + C

= ln( x + 3) - (1/2) ln(x) + C

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s.e.u.o.

Completa el trinomio cuadrado perfecto.