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Choose the correct answer. ∫1ex+e−x dx is equal to(a) tan−1ex+C(b) tan−1e−x+C (c) log (ex−e−x)+C(d) log (ex+e−x)+C |
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Answer» Choose the correct answer. ∫1ex+e−x dx is equal to(a) tan−1ex+C(b) tan−1e−x+C (c) log (ex−e−x)+C(d) log (ex+e−x)+C |
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