Can integration in which a variable is linked to (itself + constant) be solved?: $\frac{df(x)}{dx}=f(x+5)$ $$\frac{df(x)}{dx}=f(x+5)$$
I am unable to solve this kind of integration using high school mathematics.
Please help.

Just an ansatz or guess: Because differentiating $a^x$ gives a multiple of $a^x$, and also $a^{x+\mathrm{const}}$ is a multiple of $a^x$:  Try $f(x)=k\exp(ax)$, then:
$$\frac d{dx} f(x) = kae^{ax} \stackrel!= f(x+c) = ke^{a(x+c)} = ke^{ca}e^{ax}$$
with $c=5$. Dividing out $\exp(ax)$:
$$ ka \stackrel!= k\exp(ca)$$
so $k=0$ or $a = \exp(ca)$. To solve the latter, divide by $\exp(ca)$ and rewrite as
$$(-ca)\exp(-ca) = -c$$
and then apply Lambert-W:
$$-ca = W(-c) \quad \implies\quad a = -\frac1cW(-c)$$
Now for $c = 5$ there is no (real) solution because the minimum $-1/e$ of $x\mapsto xe^x$ is greater than $-5$.  But I am not familiar with $W$ and its branches, so there is likely a different branch of $W$ that gives (complex) solutions.
Or either there is no solution (except the trivial $k=0$) or the ansatz was too restricting.

That said, if you had used $c=-5$, there was the real solution
$$a = \frac15W(5) \approx1.326725/5 = 0.265345 \approx \ln 1.30388$$
so that $f(x) = k\cdot 1.30388^x$ where you can pick $k$ as you like.

Note: Here is a complex solution for $c=5$.
