integral $I=\int\dfrac{dx}{x\sqrt{2+x-x^2}}$

Question

I am trying to calculate the following integral $$I=\int\dfrac{dx}{x\sqrt{2+x-x^2}}$$

The answer in my textbook is $$-\dfrac{1}{\sqrt2}\ln\bigl|\dfrac{\sqrt{2+x-x^2}+\sqrt2}{x}+\dfrac{1}{2\sqrt2}\bigr|$$

Note that $-x^2+x+2=-(x+1)(x-2)=(2-x)(x+1),$ so the second Euler substitution is applicable. Let $$\sqrt{2+x-x^2}=t(x+1)$$ $$\sqrt{x+1}\sqrt{2-x}=t(x+1)$$ $$\sqrt{2-x}=t\sqrt{x+1} $$ $$2-t^2=x(1+t^2)$$ $$ x=\dfrac{2-t^2}{1+t^2}$$ Now we need $$\sqrt{2+x-x^2}=\dfrac{3t}{1+t^2}$$ and $$dx=\dfrac{-6t}{(1+t^2)^2}dt$$

So $$I=\int\dfrac{\frac{-6t}{(1+t^2)^2}dt}{\frac{2-t^2}{1+t^2}\cdot\frac{3t}{1+t^2}}=-2\int\dfrac{dt}{2-t^2}$$

Well, I don't see where my mistake is, but integrating the function $\dfrac{1}{(2-t^2)^2}$ won't be fun, even though it is rational.

So what am I missing, why does this problem become so technically diffucult, and what else should I try? Thanks!

Answer 1

The substitution you used is called Euler's third not second and i say mainly from personal preference it usually gives nasty rational functions

Using the actual Euler's second substitution

$$ \sqrt{2+x-x^2}=xt+\sqrt{2}\\ \ \\ x={1-2t\sqrt{2}\over1+t^2},\quad\sqrt{2+x-x^2}={-\sqrt{2}t^2+t+\sqrt{2}\over t^2+1}\\ \ \\ \mathrm dx={2\sqrt{2}t^2-2t-2\sqrt{2}\over(t^2+1)^2}\mathrm dt $$

Now the integration becomes

$$ \begin{align} I&=\int\frac{\mathrm dx}{x\sqrt{2+x-x^2}}=\int{{2\sqrt{2}t^2-2t-2\sqrt{2}\over(t^2+1)^2}\mathrm dt\over\left({1-2t\sqrt{2}\over1+t^2}\right)\left({-\sqrt{2}t^2+t+\sqrt{2}\over t^2+1}\right)}\\ &=\int{-2\over1-2\sqrt{2}t}\mathrm dt={1\over\sqrt{2}}\ln|2t\sqrt{2}-1|+C \end{align} $$

Answer 2

Now that you silently corrected your mistake (17 min after I pointed it in a comment), your "integrating the function $\dfrac{1}{(2-t^2)^2}$ won't be fun" becomes off topic and you could just go on: $$\begin{align}I&=-2\int\dfrac{dt}{2-t^2}\\ &=\frac1{\sqrt2}\ln\left|\frac{t-\sqrt2}{t+\sqrt2}\right|\\ &=\frac1{\sqrt2}\ln\left|\frac{t^2-2}{t^2+2\sqrt2t+2}\right|\\ &=-\frac1{\sqrt2}\ln\left|\frac{2-x+2\sqrt2\sqrt{2+x-x^2}+2(x+1)}{2-x-2(x+1)}\right|\\ &=-\frac1{\sqrt2}\ln\left|\frac{x+4+2\sqrt2\sqrt{2+x-x^2}}x\right|, \end{align}$$ which coincides (up to a harmless additive constant) to the answer in your textbook.

Answer 3

$$I=\int\dfrac{dx}{x\sqrt{2+x-x^2}}$$ Let $x=1/t \implies dx=-dt/t^2$ Then $$I =\int \frac{-dt}{\sqrt{2t^2+t-1}}=\frac{1}{\sqrt{2}}\int \frac{-dt}{\sqrt{t^2+t/2-1/2}}$$ $$\implies I=\frac{1}{\sqrt{2}}\int \frac{-dt}{\sqrt{(t+1/4)^2-9/16}}$$ Now try yourself from here.