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.

Let,

`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)`

Then,

`(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)`

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)`

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

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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)` .**