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$$∫(2x+10)^{-1}dx$$

$$=\frac{1}{2}∫(x+5)^{-1}dx$$

$$=\frac{1}{2}\ln|x+5|+C$$

but here is the same integral, yet different answer

$$∫(2x+10)^{-1}dx$$

Let $u = 2x+10\implies \frac{du}{dx} = 2 \implies dx = du/2$

$$\implies \frac{1}{2}∫u^{-1}du$$

$$=\frac{1}{2}\ln|u| + C$$

$$=\frac{1}{2}\ln|2x+10| + C$$

Yet,

$\frac{1}{2}\ln|2x+10|$ does not equal $\frac{1}{2}\ln|x+5|$

Please let me know if I made any mistakes. I have been seeing this over and over in my DE class and it keeps bugging me

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    $\begingroup$ $\ln(2x+10)=\ln (x+5)+\ln 2$. (So the two differ by a constant.) $\endgroup$ – David Mitra Feb 3 at 9:02
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    $\begingroup$ Welcome to MSE. Please use MathJax for formulas. $\endgroup$ – mag Feb 3 at 9:04
  • $\begingroup$ See also math.stackexchange.com/q/3495978/42969 for a similar question. $\endgroup$ – Martin R Feb 3 at 9:21
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They are still equal up to a constant (as usual with integration), since $$\log(2x+10)=\log(2(x+5))=\log(x+5)+\log2$$ by the laws of logarithms.

Another nice example of this phenomenon is $\int\sin x\cos x\,dx$, which can work out to be $-\frac14\cos2x$ (if you apply the trigonometric identity for $\sin2x$) but also $-\frac12\cos^2x$ working another way (by parts). Indeed, you should be able to find a constant $C$ such that $-\frac14\cos2x = -\frac12\cos^2x+C$ for all $x$ (by applying an appropriate trigonometric identity).


Why can you get different answers?

Remember that when you "work out" $\int f(x)\,dx$, what you're actually doing is searching for a primitive, i.e., function $F$ which satisfies $F'(x) = f(x)$. These are not uniquely determined, i.e., there are many possible functions $F$ with the property $F'=f$, but you can prove (using the mean-value theorem) that any two primitives must be equal up to a constant. (This should make sense intuitively because a constant only shifts the graph of $y=f(x)$ up and down, which does not affect the gradient of the tangent at any point).

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