This is supposed to be part of a student's Calc 2 homework; however, this seems to be an extremely difficult integration, and I couldn't figure it out.
Find the arc length of $x^3 \sqrt{9-x}$ on the interval $[0,9]$.
Progress so far:
The arc length L is given by:
$$L = \int_a^b{\sqrt{1 + {\left(\frac{\mathrm{d}y}{\mathrm{d}x}\right)}^2} \,\mathrm{d}x}$$
So we can find:
$$\frac{\mathrm{d}y}{\mathrm{d}x} = \frac{(54-7x)x^2}{2\sqrt{9-x}}$$
And eventually we wind up with this mess:
$$\int_0^9{\sqrt{1 + \frac{(54-7x)^2 x^4}{4(9-x)}} \,\mathrm{d}x}$$
Which can be expanded out to:
$$\int_0^9{\sqrt{\frac{49x^6 - 756 x^5 + 2916 x^4 - 4x + 36}{36 - 4x}} \,\mathrm{d}x}$$
The sextic polynomial in the numerator seems to be irreducible. However, it's possible to use polynomial long division on this beast, but it's unclear how that will help. For posterity after dividing the integrand is:
$$-\frac1{4}\left( 49 x^5 - 315 x^4 + 81 x^3 + 729 x^2 + 6561 x + 59045 - \frac{531441}{9-x}\right)$$
Numerically, the answer is:
$$L = 1041.550819604016\ldots$$
Does this arc have a closed form solution for its length?
How, in general, are problems like these handled?
The form you have is likely the most "closed" you will find. Arc length integrals are generally difficult because of the square root and dot product involved, and only a few can be expressed "nicely," which is why there are so few arc length exercises in textbooks. In practice, a numerical solution can be found to arbitrary precision, so we're not too concerned with a closed form.