Integration by parts MIDTERM really quick question

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banfill_89
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integration by parts! MIDTERM...really quick question...please help!

Homework Statement



Evaluate the two following integrals:
[tex]\int[/tex] xcos5x dx and [tex]\int[/tex] ln(2x+1) dx

Homework Equations





The Attempt at a Solution



ok, for the first one, the answer is 1/5 x sin5x + 1/25cos5x+C

i get this answer up until the 1/5x and the 1/25...i have no idea where there coming from. in my solutions manual they come up when u take the integral of cos5x...its 1/5sin5x...where is this 1.5 coming from?

secondly the answer to the second one is: 1/2(2x+1)ln(2x+1)-x+C

again i get this exact answer except for the 1/2. i know its coming from when u take the integral of 1/(2x+1)...but i don't get where the 1/2 comes from in front of the natural log

help would be greatly appreciated...i have a midterm wednesday!
 
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chain rule

banfill_89 said:
ok, for the first one, the answer is 1/5 x sin5x + 1/25cos5x+C

i get this answer up until the 1/5x and the 1/25...i have no idea where there coming from. in my solutions manual they come up when u take the integral of cos5x...its 1/5sin5x...where is this 1/5 coming from?

secondly the answer to the second one is: 1/2(2x+1)ln(2x+1)-x+C

again i get this exact answer except for the 1/2. i know its coming from when u take the integral of 1/(2x+1)...but i don't get where the 1/2 comes from in front of the natural log

Hi banfill_89! :smile:

It's the chain rule (backwards!) …

differentiate sin5x and you get an extra 5 because of the chain rule … same with ln(2x+1), you get an extra 2 …

so when you integrate, you have to cancel them out with a 1/5 or 1/2. :wink:
 


ohhhhhhh. makes a lot of sense...thanks a lot