Simple Calculus question, but whaaattt?

[vraeleragon]

Simple Calculus question, but whaaattt??

Hi, I have a confusing math problem. So I tried to integrate $\frac{x+1}{2}$. Apparently I got 2 answers.

answer 1. Using u substitution u=x+1, it becomes ∫$\frac{1}{2}$u du = $\frac{u^2}{4}$ --> $\frac{1}{4}$(x+1)$^{2}$
which becomes $\frac{x^2+2x+1}{4}$ = $\frac{1}{4}$x$^{2}$+$\frac{1}{2}$x+$\frac{1}{4}$

answer 2. distribute the one half. $\frac{x+1}{2}$ = $\frac{1}{2}$x+$\frac{1}{2}$
so ∫$\frac{1}{2}$x+$\frac{1}{2}$ dx = $\frac{1}{4}$x$^{2}$+$\frac{1}{2}$x

sooo, uhm, which one is right? where did the $\frac{1}{4}$ go?
thank you! and no, I'm not trolling, I'm totally serious. I have no idea how it becomes like this.

 

[vraeleragon]

oh I just remember about the C. Did the 1/4 go there?

 

[cragar]

ya that probably is it the 1/4 is just a constant. If you take the derivative of both of your answers you get the same thing. I remember running into the same thing when I integrated something with logs in it my answer didn't quite match the book and my teacher said they just expanded the log with properties of logs and got rid of the constants. If I were you I would have just puled the 1/2 out at the beginning and then integrated term by term. always pull out what ever you can.

 

[clamtrox]

You can try to figure it out by doing the integral with limits: $\int_{x_0}^x dx' \frac{x'+1}{2} = \frac{1}{2} \int_{u_0}^u du' u'$. Now $u_0$ and $x_0$ are related through $u_0 = 1 + x_0$.