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## Homework Statement

[tex]\int \frac{1}{2t^2+4} \mathrm{d}t[\tex]

## Homework Equations

[tex]\int \frac{1}{Z^2+A^2} \mathrm{d}Z = \frac{1}{A} \arctan{(\frac{Z}{A})} + c[/tex]

## The Attempt at a Solution

Looks quite easy, but this is what's annoying me: the two methods below should be identical, but something's gone wrong and I can't work out what. For some reason, the second method gives an extra factor of 1/sqrt(2) at the front. Why aren't they both the same?

[tex]\int \frac{1}{2t^2+4} \mathrm{d}t = \int \frac{1}{(\sqrt{2t})^2+2^2} \mathrm{d}t = \frac{1}{2} \arctan {(\frac{\sqrt{2}t}{2})} + c[/tex]

[tex]\int \frac{1}{2t^2+4} \mathrm{d}t = \frac{1}{2} \int \frac{1}{t^2+2} = \frac{1}{2 \sqrt{2}} \arctan{(\frac{\sqrt{2} t}{2})}+c[/tex]

With very many thanks,

Froskoy.