Another messy integral: $I=\int \frac{\sqrt{2-x-x^2}}{x^2}\ dx$

Question

I found the following question in a practice book of integration:-

$Q.$ Evaluate $$I=\int \frac{\sqrt{2-x-x^2}}{x^2}\mathrm dx$$

For this, I substituted $$t^2=\frac {2-x-x^2}{x^2}\implies x^2=\frac{2-x}{1+t^2}\implies 2t\mathrm dt=\left(-\frac4{x^3}+\frac 1{x^2}\right)\mathrm dx$$. Therefore $$\begin{align}I&=\int\frac {\sqrt{2-x-x^2}}{x^2}\mathrm dx\\&=\int \left(\frac tx\right)\left(\frac{2t\mathrm dt}{-\frac4{x^3}+\frac 1{x^2}}\right)\\&=\int \frac{2t^2\mathrm dt}{\frac{x-4}{x^2}}\\&=\int \frac{2t^2(1+t^2)\mathrm dt}{{x-4}\over{2-x}}\\&=\int \frac{2t^2(1+t^2)(5+4t^2-\sqrt{8t^2+9})\mathrm dt}{\sqrt{8t^2+9}-(8t^2+9)}\end{align}$$ Now, I substituted $8t^2+9=z^2 \implies t^2=\frac {z^2-9}8 \implies 2t\mathrm dt=\frac z4\mathrm dz$. So, after some simplification, I get $$\begin{align}I&=-\frac1{512}\int (z^2-9)(z+1)(z-1)^2\ dz\end{align}$$ I didn't have the patience to evaluate this integral after all these substitutions knowing that it can be done (I think I have made a mistake somewhere but I can't find it. There has to be an $\ln(\dots)$ term, I believe). Is there an easier way to do this integral, something that would also strike the mind quickly? I have already tried Euler substitutions but that is also messy.

Answer 1

Let $$\displaystyle I = \int \frac{\sqrt{2-x-x^2}}{x^2}dx = \int\frac{\sqrt{(x+2)(1-x)}}{x^2}dx$$

Now Let $\displaystyle (x+2) = (1-x)t^2\;,$ Then $\displaystyle x = \frac{t^2-2}{t^2+1} = 1-\frac{3}{t^2+1}$

So $$\displaystyle dx = \frac{6t}{(t^2+1)^2}dt$$ and $$\displaystyle (1-x) = \frac{3}{t^2+1}$$

So Integral $$\displaystyle I = 18\int\frac{t^2}{(t^2+1)\cdot (t^2-2)^2}dt = 6\int\frac{(t^2+1)-(t^2-2)}{(t^2+1)\cdot (t^2-2)^2}dt$$

so we get $$\displaystyle I = 6\int\frac{1}{(t^2-2)^2}dt-6\int\frac{1}{(t^2+1)(t^2-2)}dt$$

Answer 2

Hint: $\sqrt{2-x-x^2}=\sqrt{\frac94-(x+\frac12)^2}$
Substitute $x+\frac12=t; dx=dt$
$$\int \frac{\sqrt{\frac94-t^2}}{(t-\frac12)^2}\ dt$$ Put $\frac32\sin\theta=t;\frac32\cos\theta d\theta=dt$ $$\int \frac{\frac32\cos\theta.\frac32\cos\theta d\theta}{(\frac32\sin\theta-\frac12)^2}$$ $$9\int \frac{\cos^2\theta d\theta}{(9\sin^2\theta+1-6\sin \theta)}$$ Now do $u=\tan\frac{\theta}2;du=\frac12\sec^2{\frac{\theta}2}d\theta$ $$\boxed{\sin\theta=\frac{2u}{1+u^2};\cos\theta=\frac{1-u^2}{1+u^2};d\theta=\frac{2du}{u^2+1}}$$ I guess you can take it from here?

Answer 3

Let $$\displaystyle I = \int \frac{\sqrt{2-x-x^2}}{x^2}dx = \int \sqrt{2-x-x^2}\cdot \frac{1}{x^2}dx\;, $$ Now Using Integration by parts

$$\displaystyle I = -\frac{\sqrt{2-x-x^2}}{x}-\int\frac{1+2x}{2\sqrt{2-x-x^2}}\cdot \frac{1}{x}dx $$

So $$\displaystyle I = -\frac{\sqrt{2-x-x^2}}{x}-\underbrace{\int\frac{1}{\sqrt{2-x-x^2}}dx}_{J}-\underbrace{\int\frac{1}{x\sqrt{2-x-x^2}}dx}_{K}$$

So for Calculation of $$\displaystyle J = \int\frac{1}{\sqrt{2-x-x^2}}dx = \int\frac{1}{\sqrt{\left(\frac{3}{2}\right)^2-\left(\frac{2x+1}{2}\right)^2}}dx$$

Now Let $\displaystyle \left(\frac{2x+1}{2}\right)=\frac{3}{2}\sin \phi\;,$ Then $\displaystyle dx = \frac{3}{2}\cos \phi d\phi$

So we get $$\displaystyle J = \int 1d\phi = \phi+\mathcal{C_{1}} = \sin^{-1}\left(\frac{2x+1}{3}\right)+\mathcal{C}$$

Similarly for calculation of $$\displaystyle K = \int \frac{1}{x\sqrt{2-x-x^2}}dx$$

Put $\displaystyle x=\frac{1}{u}$ and $\displaystyle dx = -\frac{1}{u^2}dt$

So we get $$\displaystyle K = -\int\frac{1}{\sqrt{2u^2-u-1}}du = -\frac{1}{\sqrt{2}}\int\frac{1}{\sqrt{\left(u-\frac{1}{4}\right)^2-\left(\frac{3}{4}\right)^2}}dx$$

So we get $$\displaystyle J = -\frac{\sqrt{2}}{3}\ln\left|\left(u-\frac{1}{4}\right)+\sqrt{\left(u-\frac{1}{4}\right)^2-\left(\frac{3}{4}\right)^2}\right|+\mathcal{C_{2}}$$

So we get $$\displaystyle J = -\frac{\sqrt{2}}{3}\ln\left|\left(\frac{1}{x}-\frac{1}{4}\right)+\sqrt{\left(\frac{1}{x}-\frac{1}{4}\right)^2-\left(\frac{3}{4}\right)^2}\right|+\mathcal{C_{2}}$$

So $$\displaystyle I = -\frac{\sqrt{2-x-x^2}}{x}-\sin^{-1}\left(\frac{2x+1}{3}\right)+\frac{\sqrt{2}}{3}\ln\left|\left(\frac{1}{x}-\frac{1}{4}\right)+\sqrt{\left(\frac{1}{x}-\frac{1}{4}\right)^2-\left(\frac{3}{4}\right)^2}\right|+\mathcal{C}$$

Answer 4

WLOG $x>0$. Substitute $x=\frac1t$, then rationalize the numerator and split it as follows:

$$\begin{align}\int\frac{\sqrt{2-x-x^2}}x\mathrm dx&=\int\frac{\sqrt{2t^2-t-1}}{t}\mathrm dt\\&=\int\frac{-\frac{t}2(4t-1)-\frac{3t}2+1}{t\sqrt{2t^2-t-1}}\mathrm dt\\&=-\frac12\int\frac{\mathrm d(2t^2-t-1)}{\sqrt{2t^2-t-1}}-\frac32\int\frac{\mathrm dt}{\sqrt{2t^2-t-1}}-\int\frac{\mathrm d\left(\frac1t\right)}{\sqrt{2-\frac1t-\frac1{t^2}}}\end{align}$$

Similarly, the integral can be evaluated for $x<0$ by using $\sqrt{x^2}=-x$ while substituting $t=\frac1x$.

Answer 5

use so so-called Euler substitution and set $$\sqrt{2-x-x^2}=xt+\sqrt{2}$$

Answer 6

Standard way of doing any integrals is via riemann integration, it only requires that f(x) in $\int_a^b f(x) dx$ can be evaluated for all x\in R.

But there's even more cunning plan:

1. Calculate a list of numbers [f(0),f(1),f(2),f(3),...,f(21)]
2. Integrate using $f'(x) = \lim_{dx\rightarrow 0} \frac{f(x+dx)-f(x)}{dx}$ with dx=1 to get $f(x+1) = f(x+dx) = f(x) + dx*f'(x) = f(x)+f'(x)$.
3. Result of this magic, you get another set of numbers: $[C,\int f(0),\int f(1),\int f(2),... \int f(21)]$
4. Final step is to convert from the sample points back to a function when $((0,\int f(0)), (1,\int f(1)) ... (21,\int f(21)))$ is known.