∫1(16x2−2)dx\int\frac{1}{\left(16x^2-2\right)}dx∫(16x2−2)1dx
g−1∣f(−π4)∣g^{-1}\left|f\left(-\frac{\pi}{4}\right)\right|g−1∣∣f(−4π)∣∣
3x3−x2+x−1x+1\frac{3x^3-x^2+x-1}{x+1}x+13x3−x2+x−1
∫x2sin10xdx\int x^2\sin10xdx∫x2sin10xdx
limx→π2((π2−x)1cosx)\lim_{x\to\frac{\pi}{2}}\left(\left(\frac{\pi}{2}-x\right)\frac{1}{cosx}\right)x→2πlim((2π−x)cosx1)
cos(x)1+cos(x)−cos(x)1−cos(x)=−2cot(x)2\frac{cos\left(x\right)}{1+cos\left(x\right)}-\frac{cos\left(x\right)}{1-cos\left(x\right)}=-2cot\left(x\right)^21+cos(x)cos(x)−1−cos(x)cos(x)=−2cot(x)2
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