How do I find the antiderivative of #e^(2x) + 1#?

Answer 1
I would use the idea of integral (indefinite) and the techniques connected with this procedure: 1) I can write: #inte^(2x)+1dx# 2) I can use the fact the the integral of a sum is equal to the sum of the integrals, giving: #inte^(2x)dx+int1dx# 3) I can use the fact that the integral of the exponential is equal to itself (but here we have to consider the exponent #2x# as well) and that #1# can be written as #x^0#; #inte^(2x)dx+intx^0dx=# #e^(2x)/2+x+c#
I also evaluate the integral of #x^0# using the fact that the integral of #x^n# is #x^(n+1)/(n+1)#
4) You can now check the result (the anti-derivative) obtained above deriving it to see if it gives the initial function #e^(2x)+1#.
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Answer 2

To find the antiderivative of e^(2x) + 1, you can use the power rule for integration. The antiderivative of e^(2x) is (1/2)e^(2x), and the antiderivative of 1 is x. Therefore, the antiderivative of e^(2x) + 1 is (1/2)e^(2x) + x + C, where C is the constant of integration.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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