In the following case, suppose $b$ is a number and $b≠0$. Find
$$ \int{1\over(x^2+b^2)^n}\mathrm{d}x$$
The textbook uses the following substitution:
$$ x=bz$$ $$ \mathrm{d}x=b\mathrm{d}z$$
And reduces the integral to:
$$ \int{1\over(x^2+1)^n}\mathrm{d}x$$
However, given the presence of the exponent $n$ in the denominator and assuming that $n > 1$, how is it possible that the integral is reduced in the manner above?
If $b=0 $ we have nothing to do (the integral depend only by $x $.). Assuming $b\neq0 $, we have $$\int\frac{1}{\left(x^{2}+b^{2}\right)^{n}}dx\overset{x=bz}{=}\int\frac{b}{\left(b^{2}z^{2}+b^{2}\right)^{n}}dz=\frac{1}{b^{2n-1}}\int\frac{1}{\left(z^{2}+1\right)^{n}}dz $$ so we have the constant $1/b^{2n-1} $ outside the integral.