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how do you integrate ∫7^2x+3 dx ?
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is it: \[\int\limits_{ }^{}7^{(2x+3)}dx\] ?
7^(2x + 3) = e^ln(7^(2x+3)) = e^(2x+3)*ln(7) so integral is 7^(2x+3)/[ln(7)(2x+3)]
2x+3 =t 2dx = dt u get \[\int\limits_{}^{}7^{t}dt/2\] = 7^(t)ln 7 = 7^(2x+3) (ln7)
wait its 7^(2x+3)/(2ln7)
sorry
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7^(2x + 3) = e^ln(7^(2x+3)) = e^(2x+3)*ln(7) integral of this is e^(ln(7)*(2x+3))/[2*ln(7)] which is 7^(2x+3)/ln(49)
ys
did u understand that?
where is the guy who asked this y doesnt he respond
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