$\def\D{\displaystyle}\def\frac{\dfrac}$
(I) Let $C=\D\int \frac{\cos x}{\cos x+\sin x} d x, \quad S=\D\int \frac{\sin x}{\sin x+\cos x} d x$
\begin{eqnarray*}
C+S&=&\D\int \frac{\cos x+\sin x}{\cos x+\sin x} d x=\D\int 1 d x=x\\
C-S&=&\D\int \frac{\cos x-\sin x}{\sin x+\cos x} d x=\ln (\sin x+\cos x)\\
\therefore C&=&\frac{1}{2}(x+\ln (\sin x+\cos x))\\
S&=&\frac{1}{2}(x-\ln (\sin x+\cos x))
\end{eqnarray*}
(II) Let $C=\D\int \frac{\cos x}{\cos x-\sin x} d x, \quad S=\D\int \frac{\sin x}{\sin x-\cos x} d x$
\begin{eqnarray*}
C+S &=& \D\int \frac{\cos x-\sin x}{\cos x-\sin x} d x=\D\int 1 d x=x\\
C-S &=& \D\int \frac{\cos x+\sin x}{-\sin x+\cos x} d x=-\ln (\sin x-\cos x)\\
\therefore C &=& \frac{1}{2}(x-\ln (\sin x-\cos x))\\
S &=& \frac{1}{2}(x+\ln (\sin x-\cos x))\end{eqnarray*}
(III) Let $C=\D\int \cos x e^{x} d x ;\quad S=\D\int \sin x e^{x} d x$
\begin{eqnarray*}
\frac{d}{d x}\left(\sin x \cdot e^{x}\right)&=&(\cos x) e^{x}+\sin x\left(e^{x}\right)\\
\text { Thus } C+S&=&(\sin x) e^{x}\\
\frac{d}{d x}\left(\cos x \cdot e^{x}\right)&=&(-\sin x) e^{x}+\cos x\left(e^{x}\right)\\
\text { Thus }-S+C&=&(\cos x) e^{x}\\
\therefore \quad C&=&\frac{1}{2} e^{x}(\sin x+\cos x)\\
S&=&\frac{1}{2} e^{x}(\sin x-\cos x)
\end{eqnarray*}
(IV) Let $I=\D\int \frac{x^{2}}{x^{4}+1} d x,\quad J=\D\int \frac{1}{x^{4}+1} d x $
\begin{eqnarray*}
I+J &=& \D\int \frac{x^{2}+1}{x^{4}+1} d x=\D\int \frac{1+1 / x^{2}}{x^{2}+1 / x^{2}} d x=\D\int \frac{1+1 / x^{2}}{\left(x-\frac{1}{x}\right)^{2}+2} d x \\
&=& \D\int \frac{d u}{u^{2}+2}, \text { where } u=x-\frac{1}{x}, x \neq 0 \\
&=& \frac{1}{\sqrt{2}} \tan ^{-1}\left(\frac{u}{\sqrt 2}\right)=\frac{1}{\sqrt{2}} \tan ^{-1}\left(\frac{1}{\sqrt 2}\left(x-\frac{1}{x}\right)\right) \\
I-J &=& \D\int \frac{x^{2}-1}{x^{4}+1} d x=\D\int \frac{1-1 / x^{2}}{x^{2}+\frac{1}{x^{2}}} d x \\
&=& \D\int \frac{\left(1-1 / x^{2}\right) d x}{\left(x+\frac{1}{x}\right)^{2}-2}=\D\int \frac{d u}{u^{2}-2}, u=x+\frac{1}{x}, x \neq 0 \\
&=& \frac{1}{2 \sqrt{2}} \ln \cdot \frac{u-\sqrt{2}}{u+\sqrt{2}}=\frac{1}{2 \sqrt{2}} \ln \frac{x+\frac{1}{x}-\sqrt{2}}{x+\frac{1}{x}+\sqrt{2}} \\
\therefore \quad I &=& \frac{1}{4\sqrt{2}}\left[2 \tan ^{-1}\left(\frac{1}{\sqrt 2}\left(x-\frac{1}{x}\right)\right)+\ln \left(\frac{x+\frac{1}{x}-\sqrt{2}}{x+\frac{1}{x}+\sqrt{2}}\right)\right] \\
J &=& \frac{1}{4\sqrt{2}}\left[2 \tan ^{-1}\left(\frac{1}{\sqrt 2}\left(x-\frac{1}{x}\right)\right)-\ln \left(\frac{x+\frac{1}{x}-\sqrt{2}}{x+\frac{1}{x}+\sqrt{2}}\right)\right]
\end{eqnarray*}
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