Find the derivative of the following integral

Question

find the derivative of 

 

g(x)= ∫(u+4/u-5)du with upper bound 3x and lower bound 9x

g'(x)=?

Answer 1

Hello Dalia,

 

I saw two of this problems posted, and thought to investigate. I hope I interpret the fractions and parenthesis correct, otherwise, let me know.

 

g(x) = ∫ [(u+4)/(u-5)]du     with upper bound 3x and lower bound 9x

 

g(x) = ∫[u/(u - 5)]du + 4∫[1/(u - 5)]du

       

The second integral is   ∫[1/(u - 5)]du = ln(u - 5)

 

For the first integral let y = u - 5 or u = y+ 5 and du = dy

 

∫[u/(u - 5)]du = ∫[(y + 5)/y]dy

                     = ∫(1 + 5/y)dy

                     =  y + 5ln(y)

                     = (u - 5) + 5ln(u - 5)

Combining the two integrals 

 

g(x) = (first integral) + 4(second integral)

       =(u -5) + 5ln(u-5) + 4ln(u - 5)  

g(x) = u - 5 + 9ln(u - 5)     evaluate this at 3x and 9x

 

g(x) = [3x-5 + 9ln(3x -5)] - [9x - 5 +9ln(9x-5)]

       =  3x - 5 + 9ln(3x - 5) - 9x + 5 - 9ln(9x-5)

 

g(x)  = 9[ln(3x -5) -ln(9x - 5)] - 6x

 

g^'(x) = 9[(1/(3x-5))*3 - (1/(9x -5))*9] - 6

 

        = 27[1/(3x - 5) - 3/(9x - 5)] -6

 

g^'(x)  = 270/[(3x - 5)(9x - 5)] - 6