# How to calculate the indefinite integral of $(x-1)^{\frac 12}-(x-3)$?

Why is the following integral wrong?

$$\int[(x-1)^{\frac 12}-(x-3)]dx=\frac 23 (x-1)^{\frac32}-\frac12(x-3)^2+c$$

The answer given by my textbook is: $$\int[(x-1)^{\frac 12}-(x-3)]dx=\frac 23 (x-1)^{\frac32}-\frac12x^2+3x+c$$

I understand why the textbook's answer is right. But I don't understand what I did wrong in mine.

Note that \begin{align} \frac{2}{3}(x-1)^{3/2}-\frac{1}{2}(x-3)^2+c &= \frac{2}{3}(x-1)^{3/2}-\frac{1}{2}(x^2-6x+9)+c\\ &= \frac{2}{3}(x-1)^{3/2}-\frac{1}{2}x^2+3x\underbrace{-\frac{9}{2}+c}_{\text{constant}}, \end{align} so both answers are valid.