Suppose I have to compute the expectation of the random variable $Y=f(X)=a+bX$ with $X\sim\mathcal{N}(\mu,\sigma^2)$.
Now apparently, there are (at least) two ways of doing this. On one hand, $Y\sim\mathcal{N}(\mu+a,b^2\sigma^2)$, so \begin{equation}\tag{1} \mathbb{E}\left[Y\right]=\int_{-\infty}^{+\infty}y\;\mathcal{N}(\mu+a,b^2\sigma^2)\;\mathrm{d}y \end{equation} On the other hand, by LOTUS \begin{equation}\tag{2} \mathbb{E}[f(X)]=\int_{-\infty}^{+\infty}(a+bx)\,\mathcal{N}(\mu,\sigma^2)\mathrm{d}x \end{equation} For consistency, $(1)$ has to be equal to $(2)$, but is it? The differential is the same, apart from a $b$ factor overall.
You would get the mean of $a+bX$ using linearity of expectation. No integration required.
The distribution of Y has mean $a+b \mu$, but your first method requires you to already know that knowledge, so it’s kinda cheating going about this the wrong way.
Writing out the pdf and substituting $y$ for $a+bx$ would show your two integrals (after fixing the typo of a missing b) are the same.