$\int(x+1)dx$ yielding different results with $u$-substitution and termwise integration

Solution:

They are equivalent, they differ by a constant.

Change your second $C$ to $D$ and we have $C=\frac12 +D$.


The two results differ by a constant, which is zero when differentiated.$$\frac 12(x+1)^2=\frac 12x^2+x+\color{red}{\frac 12}=\frac 12x^2+x+\color{red}{C}$$

In fact, generally when you're evaluating an indefinite integral and you get two different results, most of the time, they're both valuable answers because they differ by a constant and not because you messed up in your work.

Of course, you can still make a mistake when evaluating indefinite integrals. I'm just saying, most of the time, it's the constant that changes the result and not your "error" you made.


Note that $F^{\prime}(x)=f(x)$ and $G^{\prime}(x)=f(x),$ then $F(x)=G(x)+C$

By Mean Value Theorem:-

If $F^{\prime}=G^{\prime}$ then $(F-G)^{\prime}=F^{\prime}-G^{\prime}=f-f=0$

Note that if a function's derivative is zero then it's a constant. So, $G(x)-F(x)=c$. $G(x)=F(x)+C$