Calculus proof: $0=1$ What is my mistake?

Question

The quotient rule states that: $$\frac{d}{dx}\frac{f(x)}{g(x)}=\frac{f'(x)g(x)-f(x)g'(x)}{{g(x)}^2}=\frac{f'(x)}{g(x)}-\frac{f(x)g'(x)}{{g(x)}^2}$$ Integrating tells us that $$\frac{f(x)}{g(x)}=\int\frac{f'(x)}{g(x)}dx-\int\frac{f(x)g'(x)}{{g(x)}^2}dx$$ If I let $f(x)=g(x)$,this gives us: $$\frac{f(x)}{f(x)}=\int\frac{f'(x)}{f(x)}dx-\int\frac{f(x)f'(x)}{{f(x)}^2}dx=\int\frac{f'(x)}{f(x)}dx-\int\frac{f'(x)}{f(x)}dx$$ Simplifying both sides: $$1=0$$ Where have I erred? Where and what is my mistake? Any help would be appreciated greatly.

Answer 1

Well, straight ahead letting $f=g$ means you're differentiating $1$ so you get $0$, but let's go through this again: $$\frac{d}{dx}\frac{f(x)}{f(x)}=\frac{f'(x)f(x)-f(x)f'(x)}{{f(x)}^2}=\frac{f'(x)}{f(x)}-\frac{f'(x)}{f(x)}$$ $$\implies 1= \int(\frac{f'(x)}{f(x)}-\frac{f'(x)}{f(x)}) dx=\int0 \ dx=C$$ So no contradiction.

Update: you did it in another way: $$ 1= \int\frac{f'(x)}{f(x)}dx-\int\frac{f'(x)}{f(x)} dx$$ Now let $y=f(x)$, we get $$\int \frac{f'(x)}{f(x)}dx= \int \frac{1}{y}dy=\ln(y)+C=\ln(f(x))+C$$ So $$1=\ln(f(x))+C_1 - (\ln(f(x))+C_2)=C_1-C_2$$ again, no contradiction.

Answer 2

When you integrated both sides, you left out the constant.

Answer 3

When integrating, you will still get a constant term. In this case the constant term is 1

Answer 4

If $ g $ and $ h $ are both differentiable at an intervall $ I $ satisfying

$$(\forall x\in I)\;\;\; \frac{d}{dx}g(x)=\frac{d}{dx}h(x)$$

THEN, there exists a Constant $ C $ such that $$(\forall x\in I)\;\; g(x)=h(x)+C$$

with $C=g(a)-h(a)$ for any $ a\in I$.

in your case, $g(x)=1$ and $h(x)=0$.