Lời giải:
Báo saiTa có:
\( \smallint {a^x}dx = \frac{{{a^x}}}{{\ln a}} + C(0 < a \ne 1)\)
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ZUNIA9
Ta có:
\( \smallint {a^x}dx = \frac{{{a^x}}}{{\ln a}} + C(0 < a \ne 1)\)