Different answer using u-sub and factoring

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Randyyy

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Hey, I had an easy integral to solve, namely: [MATH]\int \dfrac{1}{2x+1}dx[/MATH] so instead of using u-sub I thought that I´d be a little fancy and factor out a half and then bring the constant out to the front and i´d be done. It went something like this:[MATH]\int \dfrac{1}{2x+1}dx = \dfrac{1}{2} \int \dfrac{1}{x+\dfrac{1}{2}}dx = \dfrac{ln\mid x+\dfrac{1}{2} \mid }{2}+C[/MATH] But the answer is [MATH]\dfrac{ln\mid 2x+1 \mid }{2}+C[/MATH] But I don´t get why the answer would differ, I initially thought maybe you could rewrite them to one another but I really can´t see how they are the same but at the same time, I can´t imagine that my solution is incorrect.

After Closer inspection I took the derivative of both and it is the same. Am I missing something painfully obvious?
 
Randyyy said:
Hey, I had an easy integral to solve, namely: [MATH]\int \dfrac{1}{2x+1}dx[/MATH] so instead of using u-sub I thought that I´d be a little fancy and factor out a half and then bring the constant out to the front and i´d be done. It went something like this:[MATH]\int \dfrac{1}{2x+1}dx = \dfrac{1}{2} \int \dfrac{1}{x+\dfrac{1}{2}}dx = \dfrac{ln\mid x+\dfrac{1}{2} \mid }{2}+C[/MATH] But the answer is [MATH]\dfrac{ln\mid 2x+1 \mid }{2}+C[/MATH] But I don´t get why the answer would differ, I initially thought maybe you could rewrite them to one another but I really can´t see how they are the same but at the same time, I can´t imagine that my solution is incorrect.

After Closer inspection I took the derivative of both and it is the same. Am I missing something painfully obvious?
Click to expand...
Yes - I am going to be mean!

Why are you assuming that the two constants of integration - C and C - are equal????
 
So if C (in my answer) is ln(2)/2 and C (for the other answer) is 0 they are equal. But considering their derivatives are identical surely there´s a connection that I still am not seeing? I assumed the constant didn´t make any difference because if I derive them I should get back the same function and the constant = 0 when you take the derivative, so I didn´t assume they were identical, I assumed they were both insignificant but that might be where my error lies judging from your question.
 
the difference in the two forms is absorbed by the integration constant
 
Why are you assuming that the C for the other answer is 0??

How do you check to see if your answer to an integral is correct? You compute derivative of your answer. Now if you get one answer as 2(x+2) + C and another answer as 2x+ C do they have the same derivative?

Note that 3 + C, C-2 and C are all constants.
 
Jomo said:
Why are you assuming that the C for the other answer is 0??

How do you check to see if your answer to an integral is correct? You compute derivative of your answer. Now if you get one answer as 2(x+2) + C and another answer as 2x+ C do they have the same derivative?

Note that 3 + C, C-2 and C are all constants.
Click to expand...

Yes, I agree with you. I did take the derivative of both and they are identical. I also get what Romsek is saying and you alike that I could fuse multiple constants into my integration of constant. so If I had -10+e^pi+C after I was finished I could have fused them all as being the constant B. All good, My only confusion is how those two integrals produce the same derivative but they aren´t the same. I get that they could be if you factor out for example ln(2)/2 and use the props of logs and split them up and shove that into the constant. But why is this not true then for any logarithm, you could always put whatever is different into a constant and it´s the same? I´m sorry if I am being very slow and missing the bigger picture but I am struggling to grasp when this applies and when it doesn´t. Why can´t we then say that ln(5x+10)+c is the same as ln(x+9/2)+c? You could factor out the difference and put that into the constant C but their derivatives aren´t the same. However for the two expressions written above, their derivatives are exactly the same.
 
I just re-read your post. No, no, no, C is NOT (necessarily) ln(2)/2. The constant C can be any number. Repeat, the constant C can be any number.

A answer to your problem is .5ln|2x+1| + 3.4 = .5ln|x+.5| + (3.4 +ln(2)/2)
 
My only confusion is how those two integrals produce the same derivative but they aren´t the same. I get that they could be if you factor out for example ln(2)/2 and use the props of logs and split them up and shove that into the constant. This happens because the derivative of a constant is 0. You must know this if you took the derivative of both solutions and got the same answer.

ln f(x) + C = ln f(x) + (C-lnB) + lnB = ln f(x) + lnB + ln(C-lnB) = ln[Bf(x)] + ln(C-lnB) = ln[Bf(x)] + C*


Wait, you say that the integrals are not the same? So your being fancy was not correct? Where did you go wrong? Hint: your work was fine!


Why can´t we then say that ln(5x+10)+c is the same as ln(x+9/2)+c? You could factor out the difference. What are you talking about? How can ln(5x+10) differ from ln(x+9/2) by a constant. The fact that they do not give the same derivative then they can't differ by a constant. Read below.

Suppose f(x) and g(x) are differentiable and f(x) - g(x) + C. So clearly f(x) and g(x) differ by a constant. Taking the derivative gives us f'(x) - g'(x) = 0 which implies f'(x) = g'(x).
What I just showed is that if two functions differ by a constant then their derivatives are equal.

Now suppose f'(x) = g'(x). Integrating gives us that f(x) = g(x) + C or f(x) - g(x) = C. This shows that if the derivative of two functions are equal, then the two functions differ by a constant.

When we combine the two facts just proven we get that f(x) - g(x) = C if and only if f'(x) = g'(x).

All constants in an integral can be fused into one constant
 
I just tried to evaluate them both as the integral from a to b where a = 1 and b = 2 and it gave the exact same answer). It´s crazy that I never really though about it until now and integrals isn´t anything new to me. I tried removing and adding a factor of 2 in the integrals and did the exact same experiment, same derivative because they only differ by a constant and the EXACT same answer if we evaluate F(B)-F(A) which makes this so much more clear because I thought that it would become something different because they differ by a constant and they actually do not. Is there any way except taking the derivative for me to realize ahaaa, it´s the same answer that differs by a constant or is that just something you either see instantly or you have to take the derivative of both expressions to confirm?
 
[MATH]\int \dfrac{1}{2x+1}dx = \dfrac{1}{2} \int \dfrac{1}{x+\dfrac{1}{2}}dx = \dfrac{ln\mid x+\dfrac{1}{2} \mid }{2}+C_1[/MATH]
But the answer is [MATH]\dfrac{ln\mid 2x+1 \mid }{2}+C_2 [/MATH]
=ln(2∗(x+12))2+C2\displaystyle = \frac{ln \left(2 * (x + \frac{1}{2})\right)}{2} + C_2

=ln(x+12)+ln(2)2+C2\displaystyle = \frac{ln (x + \frac{1}{2}) + ln(2)}{2} + C_2

=ln(x+12)2+[ln(2)2+C2]\displaystyle = \frac{ln (x + \frac{1}{2})}{2} +\left[ \frac{ ln(2)}{2}+ C_2 \right]

=ln(x+12)2+C1\displaystyle = \frac{ln (x + \frac{1}{2})}{2} +C_1.......................where C1=[ln(2)2+C2]\displaystyle C_1 = \left[ \frac{ ln(2)}{2}+ C_2 \right]
 
I may just be replicating the great Khan here. I am going to avoid the complication of absolute value by restricting x to values > - 0.5.

[MATH]f(x) =\dfrac{1}{2} * ln(2x + 1) + k \implies[/MATH]
[MATH]f(x) =\dfrac{1}{2} * ln \left \{ 2 \left ( x + \dfrac{1}{2} \right ) \right \} + k \implies[/MATH]
[MATH]f(x) =\dfrac{1}{2} * \left \{ ln(2) + ln \left ( x + \dfrac{1}{2} \right ) \right \} + k \implies[/MATH]
[MATH]f(x) = \dfrac{ln(2)}{2} + \dfrac{1}{2} * ln \left ( x + \dfrac{1}{2} \right ) + k \implies[/MATH]
[MATH]f(x) = \dfrac{1}{2} * ln \left ( x + \dfrac{1}{2} \right ) + \dfrac{2k + ln(2)}{2}.[/MATH]
[MATH]\text {Let } K = \dfrac{2k + ln(2)}{2}.[/MATH]
[MATH]\therefore f(x) = \dfrac{1}{2} * ln \left ( x + \dfrac{1}{2} \right ) + K.[/MATH]
Just two different ways to express the exact same function.
 
Randyyy said:
I just tried to evaluate them both as the integral from a to b where a = 1 and b = 2 and it gave the exact same answer). It´s crazy that I never really though about it until now and integrals isn´t anything new to me. I tried removing and adding a factor of 2 in the integrals and did the exact same experiment, same derivative because they only differ by a constant and the EXACT same answer if we evaluate F(B)-F(A) which makes this so much more clear because I thought that it would become something different because they differ by a constant and they actually do not. Is there any way except taking the derivative for me to realize ahaaa, it´s the same answer that differs by a constant or is that just something you either see instantly or you have to take the derivative of both expressions to confirm?
Click to expand...
Yes, they will give you the same answer regardless of what you use for C.
Let's look at what you did in general.

Suppose ∫f(x)dx=F(x)+C. So ∫abf(x)dx=[F(x)+C] ∣ab=[F(b)+C]−[F(a)+C]=F(b)−F(a)\int f(x) dx = F(x) + C.\ So\ \int_a ^b f(x)dx = [F(x) + C]\ |_a^b = [F(b) + C] - [ F(a) + C ] = F(b) - F(a) . The ARBITRARY constant canceled out! Notice that if you had ∫f(x)dx=F(x)+7+C\int f(x) dx = F(x) + 7+ C that the 7 + C would cancel out in a definite integral. That is if I used the antiderivative of F(x) + 7 + C and you used the antiderivative of F(x) + C when we evaluate from a to b we would get the same answer.
 
Yes, it all makes sense Jomo! I initially thought that if we had a numerical integral the answer would be different because of the constant but of course if we had F(b)-F(a) the constants would cancel so they wouldn´t affect the answer. It somehow just looked so bizarre at first to think that they are the same but like you´ve shown above, you could easily factor out ln(2)/2 and absorb c+ln(2)/2 into a new constant say K like JeffM displayed above. Now I truly understand your first post Subhotosh Khan where you simply asked why I assumed the constant for the two expressions were the same.

Thanks for the patience and great explanations from all of you. I really appreciate it.
 
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