The value of $\int_2^4 \frac{x}{x^2 + 1} dx$ is equal to

The integral to evaluate is $\int_2^4 \frac{x}{x^2 + 1} dx$

The numerator $x$ is related to the derivative of the denominator $x^2 + 1$, suggesting a substitution approach.

Let $u = x^2 + 1$

Then $du = 2x dx$

So $x dx = \frac{du}{2}$

The limits of integration change with the substitution:

When $x = 2$: $u = 2^2 + 1 = 5$

When $x = 4$: $u = 4^2 + 1 = 17$

Substituting into the integral:

$\int_2^4 \frac{x}{x^2 + 1} dx = \int_5^{17} \frac{1}{u} \cdot \frac{1}{2} du$

$= \frac{1}{2} \int_5^{17} \frac{1}{u} du$

$= \frac{1}{2} [\ln u]_5^{17}$

$= \frac{1}{2} (\ln 17 - \ln 5)$

$= \frac{1}{2} \ln\left(\frac{17}{5}\right)$

$= \frac{1}{2} \log_e\left(\frac{17}{5}\right)$

Therefore, the value of the integral is $\frac{1}{2}\log_e\left(\frac{17}{5}\right)$