How do you evaluate the integral of $\int e^{ln (x^{2} + 1)} d x$ $int e^(ln(x^(2)+1))dx$ from -2 to -1?

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How do you evaluate the integral of $\int e^{ln (x^{2} + 1)} d x$ $int e^(ln(x^(2)+1))dx$ from -2 to -1?

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Answer 1 · 2015-12-28T16:01:26

You have that

$\int_{- 2}^{- 1} e^{ln (x^{2} + 1)} d x = \int_{- 2}^{- 1} (x^{2} + 1) d x = {[\frac{x^{3}}{3} + x]}_{- 2}^{- 1} = \frac{10}{3}$ $int_-2^-1 e^(ln(x^(2)+1))dx=int_-2^-1 (x^2+1)dx=[x^3/3+x]_-2^-1=10/3$

We used the following property

$e^{ln a} = a$ $e^(lna)=a$

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How do you evaluate the integral of $\int e^{ln (x^{2} + 1)} d x$ $int e^(ln(x^(2)+1))dx$ from -2 to -1?

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How do you evaluate the integral of $\int e^{ln (x^{2} + 1)} d x$ $int e^(ln(x^(2)+1))dx$ from -2 to -1?