Answer
$=(x^2+1)^{5x}[\frac{10x^2}{x^2+1}+5\ln (x^2+1)]$
Work Step by Step
We are given
$$h(x)=(x^2+1)^{5x}$$
From Exercise 53 we have the idea:
$\frac{d}{dx}h(x)=u(x)^{v(x)}[\frac{v(x)u'(x)}{u(x)}+(\ln u(x)v'(x))]$
so:
$\frac{d}{dx}h(x)=(x^2+1)^{5x}[\frac{5x(2x)}{x^2+1}+5\ln (x^2+1)]$
$=(x^2+1)^{5x}[\frac{10x^2}{x^2+1}+5\ln (x^2+1)]$
