How Do Absolute Values Affect Integration?

[Dethrone]

I feel like I'm asking the weirdest questions that most people don't ask, but here it is.

Suppose we have this integral (I made it up):

$$\int \sqrt{x^4+2x^3+x^2}$$

Now, I feel most people would say the answer is simply, $\frac{1}{3}x^3+\frac{1}{2}x^2+C$. But technically, that is only true when $x<-1$ or $x>0$. Are we suppose to state all the cases?

Now let's test this on a definite integral:
$$=\int_{-2}^0 \sqrt{x^4+2x^3+x^2}$$
$$=\int_{-2}^0|x^2+x|$$

$$|x^2+x|=\begin{cases}x^2+x, & x<-1, & x>0 \\[3pt] -(x^2+x), & -1<x<0 \\ \end{cases}$$

$$=\int_{-2}^{-1} x^2+x\,dx+\int_{-1}^{0} -(x^2+x)\,dx=1$$

Is this how it is supposed to be computed?

 

[I like Serena]

Is this how it is supposed to be computed?

Yep!

If you take the way "most people" would calculate it, you'll just get the wrong answer.

 

[Dethrone]

Either my workbook never did that for indefinite integrals, or I haven't encountered any of them.

$$=\int \sqrt{x^4+2x^3+x^2}$$

Case 1: if $x<-1$ or $x>0$
$$=\int x^2+x$$

Case 2: if $-1<x<0$
$$=\int -(x^2+x)$$

Stating all these cases are necessary, right?

 

[Ackbach]

Either my workbook never did that for indefinite integrals, or I haven't encountered any of them.

$$=\int \sqrt{x^4+2x^3+x^2}$$

Case 1: if $x<-1$ or $x>0$
$$=\int x^2+x$$

Case 2: if $-1<x<0$
$$=\int -(x^2+x)$$

Stating all these cases are necessary, right?

Yep.