# How would you integrate #(x^2)(e^(x-1))#?

Here, you would have to perform some partial integration.

and the integral by parts formula is applied:

Let's now obtain the values:

Thus:

Thus:

which is equivalent to:

We can now reintroduce that into our original issue and obtain:

where we can reduce things to:

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To integrate ( (x^2)(e^{x-1}) ), you can use integration by parts. Let ( u = x^2 ) and ( dv = e^{x-1}dx ). Then, ( du = 2x dx ) and ( v = e^{x-1} ). Applying the integration by parts formula:

[ \int (x^2)(e^{x-1}) dx = x^2 e^{x-1} - \int 2x e^{x-1} dx ]

Now, integrate ( 2x e^{x-1} ) using integration by parts again:

Let ( u = 2x ) and ( dv = e^{x-1}dx ). Then, ( du = 2 dx ) and ( v = e^{x-1} ).

[ \int 2x e^{x-1} dx = 2x e^{x-1} - \int 2 e^{x-1} dx ]

[ = 2x e^{x-1} - 2e^{x-1} + C ]

Substitute this back into the original equation:

[ \int (x^2)(e^{x-1}) dx = x^2 e^{x-1} - (2x e^{x-1} - 2e^{x-1}) + C ]

[ = x^2 e^{x-1} - 2x e^{x-1} + 2e^{x-1} + C ]

Therefore, the integral of ( (x^2)(e^{x-1}) ) is ( x^2 e^{x-1} - 2x e^{x-1} + 2e^{x-1} + 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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