Question

Como posso derivar (7/6) * ln (x+5) + (5/6) * ln (x-1) + C, chegando na função inicial 2x + 3 / x² + 4x - 5 ?
Obrigado

Answer 1

y=ln x  ==>y'= x'/x =1/x

derivada de constante =zero 

(7/6) * (x+5)'/(x+5)  + (5/6) * (x-1)'/(x-1)  + c'   

(7/6) * 1'/(x+5)  + (5/6) * 1/(x-1) 

7/6(x+5) + 5/6(x-1)

= (1/6)  * [7/(x+5) +5/(x-1)]

= (1/6)  * [7/(x+5) +5/(x-1)]

=(1/6) * [7(x-1) +5(x+5)]/(x+5)*(x-1)


=(1/6) * [7x-7 +5x+25]/(x+5)*(x-1)

=(1/6) * [12x+18]/(x+5)*(x-1)

= [2x+3]/(x+5)*(x-1)

= [2x+3]/(x²-x+5x-5)

= [2x+3]/(x²+4x-5)