Evaluate the integral of (x dx) / (x^2 + 2) with lower limit of 0 and upper limit of 1

$\displaystyle \int_0^1 \dfrac{x \, dx}{x^2 + 2} = \dfrac{1}{2}\int_0^1 \dfrac{2x \, dx}{x^2 + 2}$
 

Use the logarithmic function formula
$\displaystyle \int \dfrac{du}{u} = \ln u + C$

 

Hence,
$\begin{align}
\int_0^1 \dfrac{x \, dx}{x^2 + 2} & = \frac{1}{2} \Big[ \ln (x^2 + 2) \Big]_0^1 \\
& = \frac{1}{2} \Big[ \ln (1^2 + 2) - \ln (0^2 + 2) \Big] \\
& = \frac{1}{2} \Big[ \ln 3 - \ln 2 \Big] \\
& = \frac{1}{2} \, \ln \left( \dfrac{3}{2} \right) \\
& = \ln \left( \dfrac{3}{2} \right)^{1/2} \\
& = \ln \sqrt{\dfrac{3}{2}} \\
& = 0.2027 \qquad \longleftarrow \text{ answer}
\end{align}$