Willow Pratt

2022-07-06

I have to evaluate this integral:
$\int \frac{1-x+{x}^{2}}{x\sqrt{1+x-{x}^{2}}}dx$

esperoanow

You can split this into three integrals:
${I}_{1}=\int \frac{1}{x\sqrt{1+x-{x}^{2}}}dx$
${I}_{2}=\int \frac{1}{\sqrt{1+x-{x}^{2}}}dx$
${I}_{3}=\int \frac{x}{\sqrt{1+x-{x}^{2}}}dx$
For ${I}_{1}$, substitute $t=\frac{1}{x}$ and this leads to an acrosh-type integral.
For ${I}_{2}$, this is an arcsin-type integral.
For ${I}_{3}$, you can rewrite it as
$-\frac{1}{2}\int \frac{-2x+1}{\sqrt{1+x-{x}^{2}}}dx+\frac{1}{2}{I}_{2}$

Araceli Clay

$\int \frac{1-x+{x}^{2}}{x\sqrt{1+x-{x}^{2}}}dx$
The rooted is a trinomial ($a{x}^{2}+2bx+c$), which has two real roots $\alpha$ and $\beta$, decomposing it we have: $\sqrt{a{x}^{2}+2bx+c}=\sqrt{a\left(x-\alpha \right)\left(x-\beta \right)}=\left(x-\beta \right)\sqrt{\frac{a\left(x-\alpha \right)}{x-\beta }}$. Let's say $\frac{a\left(x-\alpha \right)}{x-\beta }={t}^{2}$, and we get :
$x=\frac{a\alpha -\beta {t}^{2}}{a-{t}^{2}}$
$dx=\frac{2at\left(\alpha -\beta \right)}{\left(a-{t}^{2}{\right)}^{2}}dt$
Substituting these values in the integral, we have: $\int \frac{1-x+{x}^{2}}{x\sqrt{1+x-{x}^{2}}}dx=$
$=-\int \frac{4\left(2{t}^{4}-{t}^{2}+2\right)}{\left({t}^{2}+1{\right)}^{2}\left({t}^{2}\left(\sqrt{5}+1\right)-\sqrt{5}+1\right)}dt=$
expanding
$=\int \frac{2\sqrt{5}}{\left({t}^{2}+1{\right)}^{2}}dt$$\int \frac{1-\sqrt{5}}{{t}^{2}+1}dt$$\int \frac{4}{{t}^{2}\left(\sqrt{5}+1\right)-\sqrt{5}+1\right)}dt$
Calculating the individual integrals:
${I}_{1}=\sqrt{5}.ta{n}^{-1}\left(t\right)+\frac{\sqrt{5}t}{{t}^{2}+1}$
${I}_{2}=\left(1-\sqrt{5}\right).ta{n}^{-1}\left(t\right)$
${I}_{3}=-ln\left(\frac{2t-\sqrt{5}+1}{2t+\sqrt{5}-1}\right)$
Ultimately you have: $I={I}_{1}+{I}_{2}+{I}_{3}=ta{n}^{-1}\left(t\right)+\frac{\sqrt{5}t}{{t}^{2}+1}-ln\left(\frac{2t-\sqrt{5}+1}{2t+\sqrt{5}-1}\right)$
All that remains is to replace the value of t with x:
$t=\frac{-2x-\sqrt{5}+1}{2x-\sqrt{5}-1}$
getting the result.

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