Evaluating $\int {x\over(7x-10-x^2)^{3/2}}dx$

Question

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

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

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 evaluate 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

We first use the method of completing squares. $$ I=\int \frac{x}{\left[\frac{9}{4}-\left(x-\frac{7}{2}\right)^2\right]^{\frac{3}{2}}} d x $$ Then letting $x-\frac{7}{2}=\frac{3}{2} \sin \theta$ yields $$ \begin{aligned} I & =\frac{1}{2} \int \frac{3 \sin \theta+7}{\left(\frac{3}{2}\right)^3 \cos ^3 \theta} \cdot \frac{3}{2} \cos \theta d \theta \\ & =\frac{2}{3} \int \tan \theta \sec \theta \,d \theta+\frac{14}{9} \int \sec ^2 \theta\, d \theta \\ & =\frac{2}{3} \sec \theta+\frac{14}{9} \tan \theta+C \end{aligned} $$ Plugging back $x$ gives $$ I=\frac{2(7 x-20)}{9 \sqrt{7 x-10-x^2}}+C $$

Answer 3

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.

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

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

Answer 4

Noting that the most problematic term is $\require{cancel}(-x^2+7x-10)^{-3/2}=\left((x-2)(5-x)\right)^{-3/2}$, the substitution $x=2\cos^2t+5\sin^2t, t\in\left(0,\frac\pi2\right)$ is feasible:

$$\begin{align}\int\frac{x~\mathrm dx}{(-x^2+7x-10)^{3/2}}&=\int\frac{2\cos^2+5\sin^2t}{\cancel{27}^9\sin^{\cancel3^2}t\cos^{\cancel3^2}t}\cancel6^2\cancel{\sin t}\cancel{\cos t}\mathrm dt\\&=\frac49\int\csc^2t~\mathrm dt+\frac{10}9\int\sec^2t~\mathrm dt\\&=\frac{-4}9\sqrt{\frac{5-x}{x-2}}+\frac{10}9\sqrt{\frac{x-2}{5-x}}+C\\&=\frac{14x-40}{9\sqrt{-x^2+7x-10}}+C\end{align}$$

Alternatively, break down the numerator as follows:

$$\begin{align}\frac13\int\frac{5(x-2)+2(5-x)}{\left(-x^2+7x-10\right)^{3/2}}\mathrm dx&=\frac13\int5\left(\frac{5-x}{x-2}\right)^{-3/2}+2\left(\frac{5-x}{x-2}\right)^{-1/2}\frac{\mathrm dx}{(x-2)^2}\\&=\frac{-1}9\left(4\sqrt v-\frac{10}{\sqrt v}\right)+C&v=\frac{5-x}{x-2}\\&=\frac{14x-40}{9\sqrt{-x^2+7x-10}}+C\end{align}$$

Answer 5

A generalized solution: Substitute $y=\frac{\sqrt{ax^2+bx+c}}x$ and $z=\frac cx + \frac b2$ $$\int \frac{x}{(ax^2+bx+c)^{3/2}}dx = \frac1{\Delta}\int \frac{ydz- z dy}{y^2}= \frac{z}{\Delta y}+C $$ where $\Delta= \frac{b^2}4-ac$.

Answer 6

Without integration.

$$I=\int \frac{x}{(ax^2+bx+c)^{3/2}}\,dx$$ Guess that the solution is $$I=\frac {P_n(x)}{(ax^2+bx+c)^{1/2}}$$ Differentiate both sides an remove the common denominator to obtain $$2x=2(ax^2+bx+c)P_n'(x)-(2ax+b)P_n(x)$$ If it exist $P_n(x)$ is a polynomial of low degree

Try the smallest $$P_1(x)=\alpha+\beta\,x$$ and expand

$$x (2 a \alpha -b \beta +2)+(\alpha b-2\beta c)=0$$ gives $$2 a \alpha -b \beta +2=0 \qquad \text{and} \qquad \alpha b-2\beta c=0$$ gives $$\alpha=\frac {4c}{b^2-4ac}\qquad \text{and} \qquad \beta=\frac {2b}{b^2-4ac}$$

$$I=\int \frac{x}{(ax^2+bx+c)^{3/2}}\,dx=\frac{2}{b^2-4 a c}\,\,\frac {bx+2c}{(ax^2+bx+c)^{1/2}}$$

For a sanity check, try $$P_2(x)=\alpha+\beta\,x+\gamma x^2$$ and reapeat and you will see that $\gamma=0$ and the same result for $(\alpha,\beta)$.