\(\begin{aligned}
& \text {Let } I=\int \frac{x}{\sqrt{x+1}+\sqrt{x-1}} d x \\
& =\int \frac{x[\sqrt{x+1}-\sqrt{x-1}]}{x+1-x+1} d x \\
& =\frac{1}{2} \int x \sqrt{x+1} d x-\frac{1}{2} \int x \sqrt{x-1} d x=\frac{1}{2} I_1-\frac{1}{2} I_2
\end{aligned}\)
Now, \(I_1=\int x \sqrt{x+1} d x\)
Put \(x+1=u \Rightarrow d x=d u\)
\(\begin{aligned}
\therefore I_1 & =\int(u-1) \sqrt{u} d x=\int\left(u^{\frac{3}{2}}-u^{\frac{1}{2}}\right) d x \\
& =\frac{2}{5} u^{\frac{5}{2}}-\frac{2}{3} u^{\frac{3}{2}}+c_1=\frac{2}{5}(x+1)^{\frac{5}{2}}-\frac{2}{3}(x+1)^{\frac{3}{2}}+c_1 \\
& =2(x+1)^{\frac{3}{2}}\left[\frac{1}{5}(x+1)-\frac{1}{3}\right]+c_1 \\
& =2(x+1)^{\frac{3}{2}}\left[\frac{3 x+3-5}{15}\right]+c_1 \\
& =\frac{2(3 x-2)}{15}(x+1)^{\frac{3}{2}}+c_1
\end{aligned}\)
Again, \(I_2=\int x \sqrt{x-1} d x\)
Put \(x-1=v\)
\(\begin{aligned}
\Rightarrow \quad & d x=d v \\
\therefore \quad & I_2=\int(v+1) \sqrt{v} d v=\int\left(v^{\frac{3}{2}}+v^{\frac{1}{2}}\right) d v \\
= & \frac{2}{5} v^{\frac{5}{2}}+\frac{2}{3} v^{\frac{3}{2}}+c_2=\frac{2}{5}(x-1)^{\frac{5}{2}}+\frac{2}{3}(x-1)^{\frac{3}{2}}+c_2 \\
= & 2(x-1)^{\frac{3}{2}}\left[\frac{1}{5}(x-1)+\frac{1}{3}\right]+c_2 \\
= & 2(x-1)^{\frac{3}{2}}\left[\frac{3 x-3+5}{15}\right]+c_2 \\
= & \frac{2(3 x+2)}{15}(x-1)^{\frac{3}{2}}+c_2 \\
\therefore I= & \frac{1}{2} I_1-\frac{1}{2} I_2
\end{aligned}\)
\(\begin{gathered}
=\frac{(3 x-2)}{15}(x+1)^{\frac{3}{2}}-\frac{(3 x+2)}{15}(x-1)^{\frac{3}{2}}+c \\
\quad\left[\text {where } c=c_1+c_2\right] \\
\therefore A(x)=\frac{1}{15}(3 x-2) \text { and } B(x)=\frac{-1}{15}(3 x+2] \\
\therefore A(x)+B(x)=\frac{1}{15}(3 x-2)-\frac{1}{15}(3 x+2) \\
=\frac{-4}{15}
\end{gathered}\)