Why can't $\int_1^3 \frac{4}{(2x-3)^4} dx $ be evaluated by calculators or WolframAlpha? 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?
 A: When we reason$$\int_a^bf(x)dx=[F(x)]_a^b=F(b)-F(a),$$we assume every $g$ with $g^\prime=f$ satisfies $g(b)-g(a)=F(b)-F(a)$, so the choice of $f$'s antiderivative is irrelevant.
But in general, antiderivatives differ additively by locally constant functions on the integrand's domain, which due the discontinuity at $x=\frac32$ is in this case in two components as $[1,\,3]\setminus\left\{\frac32\right\}=\left[1,\,\frac32\right)\cup\left(\frac32,\,4\right]$. We can make this even more technical, but that's all we need for now.
A locally constant function on this domain can have different values either side of $\frac32$, say $C_-$ on the left and $C_+$ on the right. But if we add such a function to an antiderivative of $4(2x-3)^{-4}$, the difference between the antiderivatives' values at $a,\,b$ with $a<\frac32<b$ changes by $C_+-C_-$, which is in general nonzero, and so the integral is improper.
A: OP: the interval change still does explain why the integral cannot be solved on calculators
This is because the "dysfunction"/discrepancy that you are raising has nothing to do with the integration-by-substitution process.
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 $$\int\frac{4}{(2x-3)^4} \mathrm dx=\frac2{3 (3 - 2 x)^3}+C.$$
This, together with Martin's example in the second comment, shows that an integrand may have an antiderivative whilst not being integrable on the desired interval.
