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**simple integral -- but a question of methodology**

[tex]\int \frac{(x-1)}{2} dx[/tex]

First, try substitution. Let [tex]u = x - 1[/tex], and [tex]du = dx[/tex]

[tex]\int \frac{u}{2} du[/tex]

[tex]\frac{1}{2} \int u \ du[/tex]

[tex]\frac{1}{2} \times \frac{u^{2}}{2}[/tex]

[tex]\frac{1}{2} \times \frac{(x-1)^2}{2}[/tex]

[tex]\frac{(x-1)^{2}}{4} + C[/tex]

Or we could do this:

[tex]\int \frac{(x-1)}{2} dx[/tex]

[tex]\int \frac{1}{2}(x-1)dx[/tex]

[tex]\frac{1}{2} \int(x-1)dx[/tex]

[tex]\frac{1}{2} (\frac{x^2}{2} - x) + C[/tex]

[tex]\frac{1}{4} (x^2 - 2x) + C[/tex]

Why should one of them be right?

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