Reduction of carboxylic acids by LiAlH4
The $\ce{Li^+}$ ion substitutes the $\ce{OH}$ in $\ce{COOH}$ first. The $\ce{C=O}$ is then reduced by $\ce{AlH_3}$ to give an aldehyde. See below:



Another $\ce{LiAlH_4}$ then reduces the aldehyde to alcohol:


I found the pictures here.