Need help solving this indefinite integral (not homework!)

Question

Source: A question bank of "tough" problems on integrals (maybe tough for a noob like me). Started by learning integration for use in physics only, but now it's got me hooked :p

Problem: Evaluate the indefinite integral $$\int {x\,\mathrm{d}x\over(7x-10-x^2)^{3/2}}.$$

I have used all the tools in my arsenal; substitution: no viable substitution comes in mind here. I have tried factoring the quadratic but that doesn't help. I have tried to multiply-divide the denominator by $x^2$ and then substitute $x={1\over t}$ but no help. I'm actually stuck right now. Please give me a hint to solve this one. All help appreciated!

@Frank gave it a shot as well...

$$\int {x\,\mathrm{d}x\over{(-1)^{3/2}(x^2-7x+10)^{3/2}}}.$$

$$\int {x\,\mathrm{d}x\over{(i)^3(x^2-7x+10)^{3/2}}}.$$ ($i$ is the imaginary unit)

Clearly we don't get any imaginary term in the answer and there are probably no chances that we'll cancel the imaginary number. That's why I did not look forward to this method. Will go ahead and try the Euler substitution...

Edit: This question is solved but I'm still looking for a better, more faster alternative as Euler's substitution can sometimes invite a bunch of calculations.

Answer 1

Thanks @DrSonnhardGraubner for giving me the right article for the problem. I didn't know about this one.

We are going to use the third substitution of Euler here, wherein we assume that

$$\sqrt{7x-10-x^2} = (5-x)t$$ (consider factorization)

$$t = \sqrt{(x-2)\over(5-x)}$$

partially differentiate to get an expression of $\mathrm{d}x$ in terms of $\mathrm{d}t$

$$\mathrm{d}x = \frac{6t \mathrm{d}t}{(t^2+1)^2}$$

Now substitute $x$ with a function of $t$ according to the above equation and get the answer as given above in the comment by @John Chessant $$-\frac{2}{9}\cdot\frac{20-7x}{\sqrt{(2-x)(5-x)}}+C$$

Any alternates are welcome!

Answer 2

By a linear transformation of the variable, you can turn to the simpler form

$$\int\frac{ax+b}{(1-x^2)^{3/2}}dx.$$

The integral of

$$\int\frac{x}{(1-x^2)^{3/2}}dx$$ is immediate, and we can focus on the other.

Using $x=\sin\theta$,

$$\int\frac1{(1-x^2)^{3/2}}dx=\int\frac{\cos\theta}{\cos^3\theta}d\theta=\tan\theta=\frac x{\sqrt{1-x^2}}$$ and you are done.

---

After seeing the shape of these expression, we can observe the useful rule

$$\left(\frac{ax+b}{\sqrt{1-x^2}}\right)'=\frac{bx+a}{(1-x^2)^{3/2}}.$$