# What is the integral of 5(6t+e^(-t))(2t+4) with respect to t?

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#int(5(6t+e^-t)(2t+4))dt#

We can split this sum up and integrate each term individually.

Combining all of the above:

Don't forget the constant of integration. Simplifying yields

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The integral of (5(6t+e^{-t})(2t+4)) with respect to (t) is:

[ \int 5(6t+e^{-t})(2t+4) , dt = \frac{15}{2}t^2 + 20t + 10e^{-t} + C ]

where (C) is the constant of integration.

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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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