Question

If $\alpha >1$, then $\int \frac{dx}{x^2+2\alpha x+1}=$.

• A. $\frac{1}{\sqrt{1-{\alpha }^2}}{\tan }^{-1}\left(\frac{x+\alpha }{\sqrt{1-{\alpha }^2}}\right)+c$
• B. $\frac{1}{\sqrt{1-{\alpha }^2}}{\cot }^{-1}\left(\frac{x+\alpha }{\sqrt{1-{\alpha }^2}}\right)+c$
• C. $\frac{1}{2\sqrt{{\alpha }^2-1}}\log \left(\frac{x+\alpha -\sqrt{{\alpha }^2-1}}{x+\alpha +\sqrt{{\alpha }^2-1}}\right)+c$
• D. $\frac{1}{2\sqrt{{\alpha }^2-1}}\log \left(\frac{x+\alpha +\sqrt{{\alpha }^2-1}}{x+\alpha -\sqrt{{\alpha }^2-1}}\right)+c$

Answer 1

The correct option is A $\frac{1}{\sqrt{1-{\alpha }^2}}{\tan }^{-1}\left(\frac{x+\alpha }{\sqrt{1-{\alpha }^2}}\right)+c$

Given : $\alpha >1,then\int \frac{dx}{(x^2+2\alpha x+1)}$

$I=\int \frac{dx}{(x^2+2\alpha x+1)}=\int \frac{dx}{(x+\alpha {)}^2+(1-{\alpha }^2)}I=\int \frac{dx}{(x+\alpha {)}^2-(\sqrt{{\alpha }^2-1}{)}^2}I=\frac{1}{2\sqrt{{\alpha }^2-1}}\log \left|\frac{x+\alpha -\sqrt{{\alpha }^2-1}}{x+\alpha -\sqrt{{\alpha }^2-1}}\right|+C$

Hence the correct answer is $\frac{1}{2\sqrt{{\alpha }^2-1}}\log \left|\frac{x+\alpha -\sqrt{{\alpha }^2-1}}{x+\alpha -\sqrt{{\alpha }^2-1}}\right|+C$