Basic integration but two different answers?

May 2010
1
0
Greetings all

I am trying to integrade

1/(x/2 + 1) dx

I get the answer 2 ln (x/2 + 1)

my friend did it this way

1/(x/2 + 1) * (2/2) dx
2/(x+2) dx

2 ln (x+2)

What is going on??
 

mr fantastic

MHF Hall of Fame
Dec 2007
16,948
6,768
Zeitgeist
Greetings all

I am trying to integrade

1/(x/2 + 1) dx

I get the answer 2 ln (x/2 + 1)

my friend did it this way

1/(x/2 + 1) * (2/2) dx
2/(x+2) dx

2 ln (x+2)

What is going on??
What's going on is that both your answers are wrong - for two reasons. One of the reasons explains why you can get two apprently different answers.

Your answer should be 2 ln |x/2 + 1| + C.

Your friends answer should be 2 ln |x+2| + K.

C and K are arbitrary constants of integration. Your teacher will have told you many many times not to forget including it.

Note: 2 ln |x/2 + 1| + C = 2 ln |1/2 (x + 2)| + C = 2 ln (1/2) + 2 ln|x+2| + C = 2 ln|x + 2| + K where K = C + 2ln(1/2).
 

chiph588@

MHF Hall of Honor
Sep 2008
1,163
429
Champaign, Illinois
Greetings all

I am trying to integrade

1/(x/2 + 1) dx

I get the answer 2 ln (x/2 + 1)

my friend did it this way

1/(x/2 + 1) * (2/2) dx
2/(x+2) dx

2 ln (x+2)

What is going on??
Let's start with your answer:

\(\displaystyle 2\ln|\tfrac x2+1| \)+C\(\displaystyle = 2\ln|\tfrac x2+1|+2\ln(2)+C_1 = 2\ln|(\tfrac x2+1)\cdot2|+C_1 \) where the last equality come from the law \(\displaystyle \ln(x)+\ln(y)=\ln(xy) \)

Thus we see \(\displaystyle 2\ln|\tfrac x2+1|+C = 2\ln|x+2|+C_1 \)

So your confusion came from the arbitrary constant you forgot to add.