# Integrating 1/(Z^2 + A^2) - something wrong?

## 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?

$$\int \frac{1}{2t^2+4} \mathrm{d}t = \int \frac{1}{(\sqrt{2t})^2+2^2} \mathrm{d}t = \frac{1}{2} \arctan {(\frac{\sqrt{2}t}{2})} + c$$

$$\int \frac{1}{2t^2+4} \mathrm{d}t = \frac{1}{2} \int \frac{1}{t^2+2} = \frac{1}{2 \sqrt{2}} \arctan{(\frac{\sqrt{2} t}{2})}+c$$

With very many thanks,

Froskoy.

Are you sure that c represents the same value in each?

Ray Vickson
Homework Helper
Dearly Missed

## 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?

$$\int \frac{1}{2t^2+4} \mathrm{d}t = \int \frac{1}{(\sqrt{2t})^2+2^2} \mathrm{d}t = \frac{1}{2} \arctan {(\frac{\sqrt{2}t}{2})} + c$$

$$\int \frac{1}{2t^2+4} \mathrm{d}t = \frac{1}{2} \int \frac{1}{t^2+2} = \frac{1}{2 \sqrt{2}} \arctan{(\frac{\sqrt{2} t}{2})}+c$$

With very many thanks,

Froskoy.

You forgot to use $\sqrt{2} dt$ in the first integral.

RGV