微积分小题集(2)

\(\newcommand \d{\ \mathrm{d}}\)

证明 \(\lim_{x \to \infty}(\sin \sqrt{x^2+2}-\sin \sqrt {x^2+1})=0\)

\(\sin x - \sin y \le |x - y|\) 即可得证。

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\[\int \frac{x^2}{1+x^2} \d x = \int (1-\frac 1{1+x^2})\d x=x - \arctan x+ C \]

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\[\int \frac{x^2}{1-x^2} \d x = \int (-1+\frac 1{1-x^2})\d x=-x+\frac 12\ln |\frac{1+x}{1-x}|+C \]

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\[\int (2^x + 3 ^ x)^2 \d x = \int (4^x + 9^x+2\cdot 6^x) \d x=\frac{4^x}{\ln 4}+\frac{9^x}{\ln 9}+2\frac{6^x}{\ln 6}+C \]

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\[\int \sqrt{1 - \sin 2x} \d x = \]

标签：frac,int,微积分,sqrt,ln,小题,sin
来源： https://www.cnblogs.com/tuagoale/p/15558858.html