Find (d/dx) ʃ(superscript 3x + 1)(subscript 2) ln(t + 1)dt.

Expert Answers
lemjay eNotes educator| Certified Educator

To solve for `d/(dx) int_2^(3x+1) ln (t+1) dt` , use the Second Fundamental Theorem of Calculus which states that if

`F(x) = int_a^xf(t)dt `        then            `F'(x) = f(x)`

To do so, apply chain rule.


       `u= 3x+1 `                

Take the derivative of u with respect to x.

  `u'(x) = (du)/(dx) = 3`

Also, let

       `y= int_2^u ln(t+1)dt`

Take the derivative of y with respect to u using the Second Fundamental Theorem.

  `y'(u) = (dy)/(du) = ln (u+1)`


 `(dy)/(dx) = (du)/(dx) * (dy)/(du) = 3*(ln u+1) = 3ln(u+1)`

Substitute back u =3x+1 .

`(dy)/(dx) = 3ln(3x+1+1) = 3ln(3x+2)`

Hence, `d/dxint_2^(3x+1) ln(t+1)dt = 3ln(3x+2)`


embizze eNotes educator| Certified Educator

Find `d/(dx)[ int_2^(3x+1) ln(t+1)dt]` :

Let `u=3x+1,(du)/(dx)=3` and let `F(x)=int_2^(3x+1)ln(t+1)dt`

Then by the chain rule `F'(x)=(dF)/(du)*(du)/(dx)`

By the 2nd FTC `d/(du)[int_2^u ln(t+1)dt]=ln(u+1)`


`d/(dx) int_2^(3x+1)ln(t+1)dt=ln((3x+1)+1)*3=3ln(3x+2)`



lfryerda eNotes educator| Certified Educator

To evaluate the derivative of the integral `d/{dx}int_2^(3x+1) ln(t+1)dt` we can use the formula 

`d/{dx}int_a^x f(t)dt=f(x)`  

combined with the chain rule.

This means that we have

`ln(3x+1+1) cdot 3`  where the final multiplication is from the derivative of `3x+1` .

This means that the derivative becomes

`3ln(3x+2)` .