Question

Evaluate: $\int \frac{dx}{(x+1)\sqrt{-x^2+x+1}}$ using Euler's substitution.

• A. $-2{\tan }^{-1}(t-1)+C,t=\frac{\sqrt{-x^2+x+1}-1}{x}$
• B. $2\tan (t-1)+C,t=\frac{\sqrt{-x^2+x+1}-1}{x}$
• C. $2{\tan }^{-1}(t+1)+C,t=\frac{\sqrt{-x^2+x+1}-1}{x}$
• D. None of these

Answer 1

The correct option is A $-2{\tan }^{-1}(t-1)+C,t=\frac{\sqrt{-x^2+x+1}-1}{x}$

$\int \frac{dx}{(x+1)\left(\sqrt{-x^2+x+1}\right)}$

$a<0$, $c>0$ ............ Third Euler substitution

$\Rightarrow x=\frac{1-2t}{t^2+1}$

$-x^2+x+1=\frac{-{(1-2t)}^2+(t^2+1)(1-2t)+{(t^2+1)}^2}{{(t^2+1)}^2}$

$=\frac{-(1+4t^2-4t)+t^2+1-2t^3-2t+t^4+1+2t^2}{{(t^2+1)}^2}$

$=\frac{t^4-t^2-2t^3-2t+1}{{(t^2+1)}^2}$

$=\frac{t^4-2t^3-t^2-2t+1}{{(t^2+1)}^2}$

$=\frac{{(t^2-t-1)}^2}{{(t^2+1)}^2}$

$\Rightarrow \sqrt{-x^2+x+1}$

$x+1=\frac{1-2t+1+t^2}{t^2+1}$

$=\frac{{(t-1)}^2+1}{t^2+1}$

$dn=\frac{(-2)(t^2+1)-\left(2t\right)(1-2t)}{{(t^2+1)}^2}dt$

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

$=\frac{2t^2-2t-2}{{(t^2+1)}^2}dt$

$\int \frac{dx}{(x+1)\left(\sqrt{-x^2+x+1}\right)}$

$2(t^2-t-1({t^2+1}^2)dt$

$=\int \frac{{(t-1)}^2+1}{(t^2+1)}\times \frac{(t^2-t-1)}{(t^2+1)}$

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

$=-2ta{n}^{-1}(t-1)+C$

$\sqrt{-x^2+x+1}=xt+1$

$t=\frac{\sqrt{-x^2+x+1}-1}{x}$