# How do you evaluate the integral of #int e^(ln(x^(2)+1))dx# from -2 to -1?

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To evaluate the integral of ( \int e^{\ln(x^2 + 1)} , dx ) from -2 to -1, we can first simplify the expression inside the exponential function using properties of logarithms:

( \ln(x^2 + 1) = \ln((-x)^2 + 1) = \ln(x^2 + 1) )

Now, we can rewrite the integral as:

( \int e^{\ln(x^2 + 1)} , dx = \int (x^2 + 1) , dx )

Integrating ( (x^2 + 1) ) with respect to ( x ) gives:

( \int (x^2 + 1) , dx = \frac{x^3}{3} + x + C )

To evaluate the definite integral from -2 to -1, we substitute the upper limit (-1) and the lower limit (-2) into the antiderivative and subtract:

( \left[\frac{x^3}{3} + x\right]_{-2}^{-1} )

Substituting the upper limit:

( \left[\frac{(-1)^3}{3} - 1\right] )

Substituting the lower limit:

( -\frac{(-2)^3}{3} + (-2) )

Simplify both expressions:

( \left[-\frac{1}{3} - 1\right] = -\frac{4}{3} - 1 )

( -\frac{-8}{3} - 2 = \frac{8}{3} - 2 )

Now, subtracting:

( \frac{8}{3} - 2 = \frac{8}{3} - \frac{6}{3} = \frac{2}{3} )

Therefore, the value of the integral ( \int e^{\ln(x^2 + 1)} , dx ) from -2 to -1 is ( \frac{2}{3} ).

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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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