Why can’t $\int_1^3 \frac{4}{(2x-3)^4} dx$ be evaluated by calculators or WolframAlpha

calculusdefinite integralsintegration

This is the integral
$$\int_1^3 \frac{4}{(2x-3)^4} dx$$

Solving by u-substitution, it works fine.
$$\text{let}~~ u = 2x-3$$

$$\int_1^34 \cdot ( 2x-3 )^{-4}dx$$

$$\ x = \frac{u+3}{2}$$

$$\ \frac{ dx }{ du }=\frac{1 }{2}$$
$$\ dx = \frac{ 1 }{ 2 }du$$

EDIT:
$$\ u(3) = 2(3)- 3$$
$$\ u(3) = 3$$
$$\ u(1) = 2(1)- 3$$
$$\ u(1) = -1$$
$$\int_{-1}^3 4 \cdot u^-4 \cdot \frac{ 1 }{ 2 }du$$

$$\int_{-1}^3 2\cdot u^{-4}du$$

$$\frac{ 2u^{-3}}{-3 } \Bigg \vert_{-1}^{3} \$$

$$= \frac{ 2}{-3(2x-3)^3 } \Bigg \vert_{-1}^{3} \$$

$$= \frac{ 2}{ -3(2(3)-3)^3 } – \frac{ 2}{ -3(2(-1)-3)^3}$$
$$= \frac{-2}{81} – \frac{2}{375}$$
$$= \frac{-84}{125}$$

But plugging this integral into a TI-84 CE Plus throws an error "Cannot divide by 0"

Also trying online calculators,
Wolframalpha and Freemathhelp, the problem is unable to be solved.

Why can't this problem be solved on these calculators?

If $$f$$ is Riemann integrable on $$[a,b]$$ and has antiderivative $$F,$$ the second fundamental theorem of calculus guarantees that $$\int_a^b f=F(b)-F(a).$$ In your example, the boldfaced condition is not satisfied (the integral fails to converge/exist due to the singularity at $$x=1.5$$), and $$\int_1^3 \frac{4}{(2x-3)^4} \mathrm dx\ne \left[\frac2{3 (3 - 2 x)^3}\right]_1^3,$$ even though $$\int\frac{4}{(2x-3)^4} \mathrm dx=\frac2{3 (3 - 2 x)^3}+C.$$