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I've got an integral like below $$\int \frac{\mathrm{d}x}{2 + \sqrt{x}}.$$

I use a substitution like this $t = 2 + \sqrt{x}$, but an answer looks a bit different. What do you suggest?

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    What do you mean, "an answer looks a bit different"? If you get an answer, you can check it by differentiating - have you done that? If the derivative of your answer is the integrand, then your answer is correct, no matter how "different" it looks.2011-06-23
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    In addition, did you do the substitution correctly? If you set $t=2+\sqrt{x}$, then you have $dt = \frac{1}{2\sqrt{x}}\,dx$, and it seems to me that it would be hard to rewrite the integral entirely in terms of $t$...2011-06-23
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    I mean an aswer in my calculus book. There the solution looks like $2\sqrt{x} - 4\log(2 + \sqrt{x}) + C$. I've got $4 + 2\sqrt{x} - 4\log(2 + \sqrt{x}) + C$.2011-06-23
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    So the only difference between the two answers is the 4 in yours, yes? But $C$ is an **arbitrary** constant. OK?2011-06-23
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    @user12465: Those are the same. See this closely related question: http://math.stackexchange.com/questions/33187/trig-integral-int-cosx-sinx-cosx-dx2011-06-23
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    Yes, it is. BTW I used $dt = \frac{1}{2\sqrt{x}}$2011-06-23
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    @Jonas Meyer OK, but what I should do that get the answer without 4? What substitution do they do there?2011-06-23
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    @Arturo: I guess $dx=(2t-4)dt$ makes it come out nice.2011-06-23
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    @Jonas: Oh, fair enough. I should stop after 12:30am (it's 1:30am now...) (-:2011-06-23
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    @user12465: You don't have to "do" anything except recognize that those are exactly the same general antiderivatives, so that you may give the one without the $4$ as the answer for its aesthetic advantage. Some would do use a different symbol to denote the arbitrary constant, e.g., "$4+C = C_1$", but this is a matter of taste.2011-06-23
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    @user12465: When you solved a problem and the solution seems right, stop questioning your solution just because the solution in your textbook is slightly different. Try and understand your solution well enough not to doubt it.2011-06-23

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The term $4$ in your answer is a constant and therefore can/should be incorporated into the constant of integration $C$.