Question

$\int \frac{dx}{(x+1)\sqrt{x^2+4}} =$

• A. $\frac{1}{\sqrt{5}} \ln\left|\frac{\sqrt{x^2+4}-x+2}{x+1}\right|+C$
• B. $\frac{1}{\sqrt{5}} \ln\left|\frac{\sqrt{x^2+4}-x-2}{x+1}\right|+C$
• C. $\frac{1}{\sqrt{5}} \ln\left|\frac{\sqrt{x^2+4}+x-2}{x+1}\right|+C$
• D. $\frac{1}{\sqrt{5}} \ln\left|\frac{\sqrt{x^2+4}-x+4}{x+1}\right|+C$

Answer 1

$\frac{1}{\sqrt{5}} \ln\left|\frac{\sqrt{x^2+4}+x-2}{x+1}\right|+C$

Answer 2

Let the given integral be $I = \int \frac{dx}{(x+1)\sqrt{x^2+4}}$. We use the substitution $x+1 = \frac{1}{t}$. Then $x = \frac{1}{t} - 1$, and $dx = -\frac{1}{t^2} dt$. Also, $x^2+4 = \left(\frac{1}{t} - 1\right)^2 + 4 = \frac{(1-t)^2}{t^2} + 4 = \frac{1 - 2t + t^2 + 4t^2}{t^2} = \frac{5t^2 - 2t + 1}{t^2}$. So, $\sqrt{x^2+4} = \sqrt{\frac{5t^2 - 2t + 1}{t^2}} = \frac{\sqrt{5t^2 - 2t + 1}}{|t|}$.

The integral becomes $I = \int \frac{-\frac{1}{t^2} dt}{\frac{1}{t} \cdot \frac{\sqrt{5t^2 - 2t + 1}}{|t|}}$.