$$ \begin{aligned} I_1 & =\int \frac{e^x}{e^{4 x}+e^{2 x}+1} d x, I_2 \\ & =\int \frac{e^{-x}}{e^{-4 x}+e^{-2 x}+1} d x, \text { then } I_2-I_1= \end{aligned} $$
If $\int \frac{1}{x} \sqrt{\frac{1-\sqrt{x}}{1+\sqrt{x}}} d x=2 f(x)-2 \sin ^{-1} \sqrt{x}+c$, then $f(x)=$
$$ \begin{aligned} & \int \frac{3 x+2}{4 x^2+4 x+5} d x=A \log \\ & \left(4 x^2+4 x+5\right)+B \tan ^{-1}\left(\frac{2 x+1}{2}\right)+C, \text { then } A+B= \end{aligned} $$
Consider the following
Assertion
$$ \begin{aligned} & \text { (A) } \begin{array}{r} \sqrt{x-3}\left(\sin ^{-1}(\log x)+\cos ^{-1}\right. \\ (\log x) d x=\frac{\pi}{3}(x-3)^{3 / 2}+c \end{array} \end{aligned} $$
Reason $(\mathrm{R}) \sin ^{-1}(f(x))+\cos ^{-1}(f(x))=\frac{\pi}{2},|f(x)|<1$
The correct answer is
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