<s>[INST] Changing bounds of integration question; spot the mistake. Suppose we wanted to compute $$\int \frac{1}{x^2 + x + 1} ~ dx.$$ One such way is to fix $c$ and let $x$ vary as bounds, thus we have that $$\int \frac{1}{x^2 + x + 1} ~ dx = \int_c^x \frac{1}{t^2 + t + 1} ~ dt = \int_c^x \frac{1}{(t + 1/2)^2 + 3/4} ~ dt.$$ We will now shift the integral by $1/2,$ so we have $$\int_c^x \frac{1}{(t + 1/2)^2 + 3/4} ~ dt = \int_{c+1/2}^{x + 1/2} \frac{1}{t^2 + 3/4} ~ dt.$$ The expansion theorem for integrals tells us that $$\int_{a}^b f(x + k) ~ dx = \frac{1}{c} \int_{ac + k}^{bc + k} f(x/c) ~ dx.$$ Using the aforementioned theorem, we have that $$\int_{c+1/2}^{x + 1/2} \frac{1}{t^2 + 3/4} ~ dt = \frac{\sqrt{3}}{2} \int_{\frac{2}{\sqrt{3}}c+1/2}^{\frac{2}{\sqrt{3}}x + 1/2} \frac{1}{(3/4)t^2 + 3/4} ~ dt.$$ Factoring, we have $$\frac{\sqrt{3}}{2} \int_{\frac{2}{\sqrt{3}}c+1/2}^{\frac{2}{\sqrt{3}}x + 1/2} \frac{1}{(3/4)t^2 + 3/4} ~ dt = \frac{\sqrt{3}}{2}\cdot\frac{4}{3} \int_{\frac{2}{\sqrt{3}}c+1/2}^{\frac{2}{\sqrt{3}}x + 1/2} \frac{1}{t^2 + 1} ~ dt.$$ This of course evaluates to $$\frac{2\sqrt{3}}{3} \arctan\left( \frac{2}{\sqrt{3}}x + \frac{1}{2} \right) + \text{constant}.$$ But then notice, taking the derivative of the expression before will not give us $1/(x^2 + x + 1).$ But, taking the derivative of $$\frac{2\sqrt{3}}{3} \arctan\left( \frac{2}{\sqrt{3}}x + \frac{1}{\sqrt{3}} \right) + \text{constant}$$ does give us $1/(x^2 + x + 1).$ Where is the mistake?
 [/INST] $$\int_{a}^b f(x + k) ~ dx = \frac{1}{c} \int_{ac + k}^{bc + k} f(x/c) ~ dx.$$ is wrong. It should be:
$$\int_{a}^b f(x + k) ~ dx = \frac{1}{c} \int_{a(c + k)}^{b(c + k)} f(x/c) ~ dx.$$
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