$\int\limits_{\sqrt{log_e 2}}^{\sqrt{log_e 4}} xe^{x^2} dx$ is equal to

We need to find:

$\int_{\sqrt{\log_e 2}}^{\sqrt{\log_e 4}} x \, e^{x^2} \, dx$

$\log_e$ is the same thing as $\ln$ (natural log). We'll use $\ln$ from here on.

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Simplify the limits:

Lower limit $= \sqrt{\ln 2}$

Upper limit $= \sqrt{\ln 4}$

Since $4 = 2^2$, we use the log rule $\ln(a^n) = n\ln a$:

$\sqrt{\ln 4} = \sqrt{\ln(2^2)} = \sqrt{2\ln 2}$

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Look at the integrand: $x \, e^{x^2}$. There's an $x$ multiplied with $e^{x^2}$. The derivative of $x^2$ is $2x$ — and we already have that $x$ sitting right there. This is a direct hint to substitute $u = x^2$.

Let $u = x^2$

$du = 2x \, dx \implies x \, dx = \dfrac{du}{2}$

The integral becomes:

$\int x \, e^{x^2} \, dx = \int e^u \cdot \dfrac{du}{2}$

$= \dfrac{1}{2} \int e^u \, du$

$= \dfrac{1}{2} \, e^u$

$= \dfrac{1}{2} \, e^{x^2}$

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Apply the limits:

$\left[\dfrac{1}{2} \, e^{x^2}\right]_{\sqrt{\ln 2}}^{\sqrt{2\ln 2}}$

$= \dfrac{1}{2} \, e^{(\sqrt{2\ln 2})^2} - \dfrac{1}{2} \, e^{(\sqrt{\ln 2})^2}$

The square root and the square cancel each other out since $(\sqrt{a})^2 = a$:

$= \dfrac{1}{2} \, e^{2\ln 2} - \dfrac{1}{2} \, e^{\ln 2}$

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Since $e$ and $\ln$ are inverse functions, $e^{\ln(a)} = a$:

$e^{\ln 2} = 2$

For $e^{2\ln 2}$, rewrite the exponent: $2\ln 2 = \ln(2^2) = \ln 4$

$e^{2\ln 2} = e^{\ln 4} = 4$

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$= \dfrac{1}{2}(4) - \dfrac{1}{2}(2)$

$= 2 - 1$

$= 1$