Regioselectivity of bromination of alkenes
With bromination, the first step is the formation of an adduct complex, which can be explained that the &pi;-bond polarises the dibromide. The next step is the formation of the bromonium ion, which carries an overall positive charge. A symmetry-changing induction like substituents will always result in asymmetric carbon-bromine bond lengths in the bromonium ion. Since it is a cation, similar rules for the stability/instability as for carbocations apply. Therefore the higher substituted carbon will have the longer $\ce{C-Br}$ bond length, and the higher charge (more positive/less negative). This makes it a better target for the residual nucleophile $\ce{Br-}$ to attack there.

Here is the bromonium ion for the reaction
$$
\ce{\underset{alkene}{H3C-CH=CH2} + Br2 -&gt; 
[\underset{bromonium ion}{H3C-CH(Br)CH2}]+ + Br- -&gt; 
\underset{$trans-$alkane}{H3C-CHBr-CH2Br}}
$$
calculated at the DF-b97d3/def2-SVP level of theory, charges with NBO6, to illustrate the concept:



Additionally the HOMO has the higher coefficient at the lesser substituted carbon, while the LUMO has the higher coefficient at the higher substituted carbon.




From this, you can see that both quotes from the textbooks are correct.