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.
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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. $\displaystyle u=\ln(2x+3) \implies du=\frac{2}{2x+3}dx$ Can you finish from here?
Originally Posted by TheEmptySet $\displaystyle u=\ln(2x+3) \implies du=\frac{2}{2x+3}dx$ 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?
you will substitute du/2 for dx/(2x+3) and then integrate with respect to u.
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