I have to evaluate this integral:

$\int \frac{1-x+{x}^{2}}{x\sqrt{1+x-{x}^{2}}}dx$

$\int \frac{1-x+{x}^{2}}{x\sqrt{1+x-{x}^{2}}}dx$

Willow Pratt
2022-07-06
Answered

I have to evaluate this integral:

$\int \frac{1-x+{x}^{2}}{x\sqrt{1+x-{x}^{2}}}dx$

$\int \frac{1-x+{x}^{2}}{x\sqrt{1+x-{x}^{2}}}dx$

You can still ask an expert for help

esperoanow

Answered 2022-07-07
Author has **11** answers

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

$-\frac{1}{2}\int \frac{-2x+1}{\sqrt{1+x-{x}^{2}}}dx+\frac{1}{2}{I}_{2}$

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

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

${I}_{3}=\int \frac{x}{\sqrt{1+x-{x}^{2}}}dx$

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

Answered 2022-07-08
Author has **2** answers

$\int \frac{1-x+{x}^{2}}{x\sqrt{1+x-{x}^{2}}}dx$

The rooted is a trinomial ($a{x}^{2}+2bx+c$), which has two real roots $\alpha $ and $\beta $, decomposing it we have: $\sqrt{a{x}^{2}+2bx+c}=\sqrt{a(x-\alpha )(x-\beta )}=(x-\beta )\sqrt{\frac{a(x-\alpha )}{x-\beta}}$. Let's say $\frac{a(x-\alpha )}{x-\beta}={t}^{2}$, and we get :

$x=\frac{a\alpha -\beta {t}^{2}}{a-{t}^{2}}$

$dx=\frac{2at(\alpha -\beta )}{(a-{t}^{2}{)}^{2}}dt$

Substituting these values in the integral, we have: $\int \frac{1-x+{x}^{2}}{x\sqrt{1+x-{x}^{2}}}dx=$

$=-\int \frac{4(2{t}^{4}-{t}^{2}+2)}{({t}^{2}+1{)}^{2}({t}^{2}(\sqrt{5}+1)-\sqrt{5}+1)}dt=$

expanding

$=\int \frac{2\sqrt{5}}{({t}^{2}+1{)}^{2}}dt$$\int \frac{1-\sqrt{5}}{{t}^{2}+1}dt$$\int \frac{4}{{t}^{2}(\sqrt{5}+1)-\sqrt{5}+1)}dt$

Calculating the individual integrals:

${I}_{1}=\sqrt{5}.ta{n}^{-1}(t)+\frac{\sqrt{5}t}{{t}^{2}+1}$

${I}_{2}=(1-\sqrt{5}).ta{n}^{-1}(t)$

${I}_{3}=-ln(\frac{2t-\sqrt{5}+1}{2t+\sqrt{5}-1})$

Ultimately you have: $I={I}_{1}+{I}_{2}+{I}_{3}=ta{n}^{-1}(t)+\frac{\sqrt{5}t}{{t}^{2}+1}-ln(\frac{2t-\sqrt{5}+1}{2t+\sqrt{5}-1})$

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.

The rooted is a trinomial ($a{x}^{2}+2bx+c$), which has two real roots $\alpha $ and $\beta $, decomposing it we have: $\sqrt{a{x}^{2}+2bx+c}=\sqrt{a(x-\alpha )(x-\beta )}=(x-\beta )\sqrt{\frac{a(x-\alpha )}{x-\beta}}$. Let's say $\frac{a(x-\alpha )}{x-\beta}={t}^{2}$, and we get :

$x=\frac{a\alpha -\beta {t}^{2}}{a-{t}^{2}}$

$dx=\frac{2at(\alpha -\beta )}{(a-{t}^{2}{)}^{2}}dt$

Substituting these values in the integral, we have: $\int \frac{1-x+{x}^{2}}{x\sqrt{1+x-{x}^{2}}}dx=$

$=-\int \frac{4(2{t}^{4}-{t}^{2}+2)}{({t}^{2}+1{)}^{2}({t}^{2}(\sqrt{5}+1)-\sqrt{5}+1)}dt=$

expanding

$=\int \frac{2\sqrt{5}}{({t}^{2}+1{)}^{2}}dt$$\int \frac{1-\sqrt{5}}{{t}^{2}+1}dt$$\int \frac{4}{{t}^{2}(\sqrt{5}+1)-\sqrt{5}+1)}dt$

Calculating the individual integrals:

${I}_{1}=\sqrt{5}.ta{n}^{-1}(t)+\frac{\sqrt{5}t}{{t}^{2}+1}$

${I}_{2}=(1-\sqrt{5}).ta{n}^{-1}(t)$

${I}_{3}=-ln(\frac{2t-\sqrt{5}+1}{2t+\sqrt{5}-1})$

Ultimately you have: $I={I}_{1}+{I}_{2}+{I}_{3}=ta{n}^{-1}(t)+\frac{\sqrt{5}t}{{t}^{2}+1}-ln(\frac{2t-\sqrt{5}+1}{2t+\sqrt{5}-1})$

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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