# Possible counter-proof of the constant rule of integration [duplicate]

$$∫(2x+10)^{-1}dx$$

$$=\frac{1}{2}∫(x+5)^{-1}dx$$

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

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

• $\ln(2x+10)=\ln (x+5)+\ln 2$. (So the two differ by a constant.) – David Mitra Feb 3 at 9:02
• Welcome to MSE. Please use MathJax for formulas. – mag Feb 3 at 9:04
• See also math.stackexchange.com/q/3495978/42969 for a similar question. – Martin R Feb 3 at 9:21

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).
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).