I'm confused as both solutions seem to be correct but they give two different answers
O OutBoxer024 New member Joined May 5, 2024 Messages 1 May 5, 2024 #1 I'm confused as both solutions seem to be correct but they give two different answers
L lookagain Elite Member Joined Aug 22, 2010 Messages 3,258 May 7, 2024 #2 OutBoxer024 said: View attachment 37791 I'm confused as both solutions seem to be correct but they give two different answers Click to expand... In the third line down on the left-hand side, you are missing a (1/3) multiplier in front of the integral. The answer at the bottom right of the paper can be rewritten as ln∣3(x+3)∣3 + C1 =\displaystyle \dfrac{ ln|3(x + 3)|}{3} \ + \ C_1 \ = 3ln∣3(x+3)∣ + C1 = ln∣3∣ + ln∣x+3∣3 +C1 =\displaystyle \dfrac{ln|3| \ + \ ln|x + 3|}{3} \ + C_1 \ = 3ln∣3∣ + ln∣x+3∣ +C1 = ln∣3∣3 + ln∣x+3∣3 + C1 =\displaystyle \dfrac{ln|3|}{3} \ + \ \dfrac{ln|x + 3|}{3} \ + \ C_1 \ = 3ln∣3∣ + 3ln∣x+3∣ + C1 = ln∣x+3∣3 + (C1 + ln∣3∣3) =\displaystyle \dfrac{ln|x + 3|}{3} \ + \ \ \bigg(C_1 \ + \ \dfrac{ln|3|}{3}\bigg) \ = 3ln∣x+3∣ + (C1 + 3ln∣3∣) = ln∣x+3∣3 + C\displaystyle \dfrac{ln|x + 3|}{3} \ + \ C 3ln∣x+3∣ + C This is true because ln∣3∣3 \displaystyle \ \dfrac{ln|3|}{3} \ 3ln∣3∣ is a constant, and ( C1 + ln∣3∣3) \displaystyle \bigg( \ C_1 \ + \ \dfrac{ln|3|}{3} \bigg) \ ( C1 + 3ln∣3∣) is a new constant, which can be renamed as "C."
OutBoxer024 said: View attachment 37791 I'm confused as both solutions seem to be correct but they give two different answers Click to expand... In the third line down on the left-hand side, you are missing a (1/3) multiplier in front of the integral. The answer at the bottom right of the paper can be rewritten as ln∣3(x+3)∣3 + C1 =\displaystyle \dfrac{ ln|3(x + 3)|}{3} \ + \ C_1 \ = 3ln∣3(x+3)∣ + C1 = ln∣3∣ + ln∣x+3∣3 +C1 =\displaystyle \dfrac{ln|3| \ + \ ln|x + 3|}{3} \ + C_1 \ = 3ln∣3∣ + ln∣x+3∣ +C1 = ln∣3∣3 + ln∣x+3∣3 + C1 =\displaystyle \dfrac{ln|3|}{3} \ + \ \dfrac{ln|x + 3|}{3} \ + \ C_1 \ = 3ln∣3∣ + 3ln∣x+3∣ + C1 = ln∣x+3∣3 + (C1 + ln∣3∣3) =\displaystyle \dfrac{ln|x + 3|}{3} \ + \ \ \bigg(C_1 \ + \ \dfrac{ln|3|}{3}\bigg) \ = 3ln∣x+3∣ + (C1 + 3ln∣3∣) = ln∣x+3∣3 + C\displaystyle \dfrac{ln|x + 3|}{3} \ + \ C 3ln∣x+3∣ + C This is true because ln∣3∣3 \displaystyle \ \dfrac{ln|3|}{3} \ 3ln∣3∣ is a constant, and ( C1 + ln∣3∣3) \displaystyle \bigg( \ C_1 \ + \ \dfrac{ln|3|}{3} \bigg) \ ( C1 + 3ln∣3∣) is a new constant, which can be renamed as "C."
Steven G Elite Member Joined Dec 30, 2014 Messages 14,598 May 7, 2024 #3 Whenever this happens just check to see if your results differ by a constant.