∫(2^x/log2)dx=(1/log2)∫(2^x)dx (1/log2は定数なので、前に出す)
=(1/log2)×(2^x)log2 (∫{a^x}dx={a^x}logaより)
=(2^x)×log2×(1/log2)
=2^x
となります。
xが含まれていないので、定数です。
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1/log2は定数なんですか？！