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$$
\int\left(\sec ^4 x+\tan ^4 x\right) d x=
$$
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\int\left(\sec ^4 x+\tan ^4 x\right) d x=
$$
Solution:
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Verified Answer
The correct answer is:
$\frac{2}{3} \tan ^3 x+x+c$
$\begin{aligned} & \text { } I=\int\left(\sec ^4 x+\tan ^4 x\right) d x \\ & =\int\left[\sec ^4 x+\left(\sec ^2 x-1\right)^2\right] d x \\ & =\int\left(2 \sec ^4 x-2 \sec ^2 x+1\right) d x \\ & =\int\left[2\left(1+\tan ^2 x\right) \sec ^2 x-2 \sec ^2 x+1\right] d x \\ & =\int\left(2 \tan ^2 x \sec ^2 x+1\right) d x=\frac{2}{3} \tan ^3 x+x+c .\end{aligned}$
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