1. ## U substitution

This seems like it might be very obvious but I have trouble with U substitution. Can someone show me how they would solve this by U substitution?

the integral from .5 to 1 of ln(2x+3) over 2x+3 dx.

2. Originally Posted by mike21
This seems like it might be very obvious but I have trouble with U substitution. Can someone show me how they would solve this by U substitution?

the integral from .5 to 1 of ln(2x+3) over 2x+3 dx.
$u=\ln(2x+3) \implies du=\frac{2}{2x+3}dx$

Can you finish from here?

3. Originally Posted by TheEmptySet
$u=\ln(2x+3) \implies du=\frac{2}{2x+3}dx$

Can you finish from here?
so then you plug u in so it would look like u over 2x+3 but do you integrate in respect to u or x?

4. you will substitute du/2 for dx/(2x+3) and then integrate with respect to u.