Question

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

• A. $-\frac{1}{\sqrt{5}}{\sin }^{-1}\frac{7+4x}{3(2x+1)}+c$
• B. $-\frac{1}{\sqrt{5}}cos\frac{7+4x}{3(2x+1)}+c$
• C. $-\frac{1}{\sqrt{5}}sin{h}^{-1}\frac{7+4x}{3(2x+1)}+c$
• D. $-\frac{1}{\sqrt{5}}cos{h}^{-1}\frac{7+4x}{3(2x+1)}+c$

Answer 1

The correct option is A $-\frac{1}{\sqrt{5}}{\sin }^{-1}\frac{7+4x}{3(2x+1)}+c$

$I=\int \frac{1}{(2x+1)\sqrt{x^2-x-2}}dxSubstitute2x+1=\frac{1}{t}x=\frac{1-t}{2t}dx=\frac{-1}{2t^2}dt\sqrt{x^2-x-2}=\frac{\sqrt{1-5t^2-4t}}{2t}I=-\int \frac{1}{\sqrt{1-5t^2-4t}}dt=-\int \frac{1}{\sqrt{{\left(\frac{3}{\sqrt{5}}\right)}^2-{\left(\frac{5t+2}{\sqrt{5}}\right)}^2}}dt=-\frac{1}{\sqrt{5}}{\sin }^{-1}\left(\frac{5t+2}{3}\right)+c=-\frac{1}{\sqrt{5}}{\sin }^{-1}\left(\frac{7+4x}{3(2x+1)}\right)+c$