your answer for $\displaystyle \int \frac {x}{x^2 + 3}~dx$ is wrong. do it again. use substitution, $\displaystyle u = x^2 + 3$

for $\displaystyle \int \frac {1}{x^2 + 3}~dx$ we proceed as follows.

note that: $\displaystyle \frac {1}{x^2 + 3} = \frac {1}{3 \left( \frac {x^2}{3} + 1 \right)}$

$\displaystyle = \frac {1}{3} \cdot \frac {1}{ \left( \frac {x}{\sqrt {3}} \right)^2 + 1 }$

so, $\displaystyle \int \frac {1}{x^2 + 3}~dx = \frac {1}{3} \int \frac {1}{ \left( \frac {x}{\sqrt {3}} \right)^2 + 1 }~dx$

now continue