Question

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

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

Answer 1

The correct option is A $-2{\tan }^{-1}\sqrt{\frac{1-x}{x+2}}+C$

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

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

$\Rightarrow -x^2-x+2=x^2{t}^2+2+2\sqrt{2}xt\Rightarrow x=-\frac{(2\sqrt{2}t+1)}{(1+t^2)}\Rightarrow dx=\frac{(2\sqrt{2}{t}^2+2t-2\sqrt{2})dt}{{(1+t^2)}^2}$

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

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

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

$=-2\int \frac{dt}{(1+t^2)}=-2{\tan }^{-1}\left(t\right)=-2{\tan }^{-1}\sqrt{\frac{1-x}{x+2}}+C$