(a) $\frac{1}{5} \tan 5 x-x+c $;

(b)$\frac{1}{2} \sin b-\frac{1}{10} \sin 5 b$ ;

(c) $\ln \frac{\ln b}{\ln a}$;

(d) $\frac{1}{4\left(1+e^{2 x}\right)^2}+c $

(a)
$$
\begin{array}{l}
\int \tan ^2 5 x d x \\
=\int \sec ^2 5 x-1 d x \\
=\frac{1}{5} \tan 5 x-x+c
\end{array}
$$
(b)
$$
\begin{array}{l}
\int_0^b \sin 2 x \sin 3 x d x \\
=-\frac{1}{2} \int \cos 5 x-\cos x d x \\
=-\frac{1}{2}\left[\frac{1}{5} \sin 5 x-\sin x\right]_0^b \\
=-\frac{1}{2}\left[\frac{1}{5} \sin 5 b-\sin b\right]
\end{array}
$$
(c)
$$
\begin{array}{l}
\int_a^b \frac{1}{x \ln x} d x \\
=[\ln |\ln x|]_a^b \\
=\ln (\ln b)-\ln (\ln a) \text { since } 1<a<b \\
=\ln \left(\frac{\ln b}{\ln a}\right)
\end{array}
$$
(d)
$$
\begin{array}{l}
u=1+e^{2 x} \Rightarrow \frac{d u}{d x}=2 e^{2 x} \\
\int \frac{e^{2 x}}{\left(1+e^{2 x}\right)^3} d x \\
=\int \frac{e^{2 x}}{u^3} \cdot \frac{1}{2 e^{2 x}} d u \\
=\int \frac{1}{u^3} d u \\
=-\frac{1}{4 u^2}+c \\
=-\frac{1}{4\left(1+e^{2 x}\right)^2}+c
\end{array}
$$