of ∫ (e^t + 1/t)dt
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To take the integration of e^t, you have to take the dirivative of it's exponent and multiply it to the original equation.
int.] e^t = 1 * e^t (as d/dt (t) = 1)
Next, there is a formula that states the integral of 1/x is ln x.
int.] 1/t = ln (t)
Lastly, whenever you do an integral you also must take into account the the antiderivative could have had a constant (as the derivative of a constant is zero). So to account for this potential loss, you would just add C. So your answer is
e^(t) + ln (t) + C
int.] e^t = 1 * e^t (as d/dt (t) = 1)
Next, there is a formula that states the integral of 1/x is ln x.
int.] 1/t = ln (t)
Lastly, whenever you do an integral you also must take into account the the antiderivative could have had a constant (as the derivative of a constant is zero). So to account for this potential loss, you would just add C. So your answer is
e^(t) + ln (t) + C
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As given MUST be read as :-
I = ∫ e^t + (1/t) dt
I = e^t + ln t + C
I = ∫ e^t + (1/t) dt
I = e^t + ln t + C
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= e^(t) + ln|t| + C
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e^t + Log[t]