I have a doubt regarding basic integration $$ \int (1+x) \mathrm{d} x= \int 1 \mathrm{d}x+ \int x \mathrm{d} x =x+ \frac{x^2}{2} +c $$ But in another method it can also be done as $$ \int(1+x)\mathrm{d}x= \frac{(1+x)^2}{2} +c $$ Both these methods lead to different answers. Is is because of integration constant $c$ or is any method wrong, if so why is it wrong. Please forgive my stupidity.
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8Expand $(1+x^)2$ and combine $1/2$ with $c$. Also try to use mathjax, only then you can be forgiven – Hyperbolic PDE friend Mar 23 '23 at 16:47
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You could rename your integration constants to $c_1$ and $c_2$ respectively to avoid confusion. – cs89 Mar 23 '23 at 16:51
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The two answers you have are actually the same answer. If you keep expanding...
$$ \int(1+x)\mathrm{d}x= \frac{(1+x)^2}{2} +c $$
$$ = \frac{1 + 2x + x^2 }{2} + c = \frac{1}{2} + \frac{2x}{2} + \frac{x^2}{2} + c $$
$$ = \frac{1}{2} + x + \frac{x^2}{2} + c = x + \frac{x^2}{2} + \frac{1}{2} + c $$
Note that $\frac{1}{2} + c = c'$. In other words, for some arbitrary constant of integration $c$, the sum $\frac{1}{2} + c$ is just some other arbitrary constant of integration $c'$. So we have
$$ = x + \frac{x^2}{2} + c' $$
Remember an indefinite integral $F(x) + c$ is not a unique solution. It actually represents a whole set of solutions obtained by adding any fixed constant $c$ to $F(x)$ where $F(x)$ is assumed to have no terms that are fixed constants. For this reason, we can combine $\frac{1}{2} + c $ into a new constant $c'$ and say $ x + \frac{x^2}{2} + c' $ belongs to the same family of solutions as your first answer.
RyRy the Fly Guy
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The answers look different, but are actually the same. You're right that it's because of the integration constant.
The first answer is:
$$x + \frac{x^2}{2} + \color{red}{C}$$
where the integration constant means: the whole family of functions including any value for the constant $\color{red}{C}$ such as $$x + \frac{x^2}{2} + \color{red}{1},\quad x + \frac{x^2}{2} + \color{red}{-3}, \ldots$$
The second answer is:
$$\frac{(1+x)^2}{2} + \color{skyblue}{C} = x + \frac{x^2}{2} + \frac{1}{2} + \color{skyblue}{C}$$
The key idea is that any function you can make by choosing a value of $\color{red}{C}$ in $x + \frac{x^2}{2} + \color{red}{C}$, you can also make by choosing an appropriate value of $\color{skyblue}{C}$ in $\frac{(1+x)^2}{2} + \color{skyblue}{C} = x + \frac{x^2}{2} + \frac{1}{2} + \color{skyblue}{C}$.
This is because these two expressions are the same, except for a constant difference of $\frac{1}{2}$.
user326210
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