Question

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

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

Answer 1

The correct option is B $\frac{1}{2}[\sqrt{x^2+2x+2}-x+\ln |\sqrt{x^2+2x+2}-x-2|+\ln |\frac{\sqrt{x^2+2x+2}-x-2}{\sqrt{x^2+2x+2}-x-1}|]+C$

$Q\int \frac{1}{x+\sqrt{x^2+2x+2}}dx$

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

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

$\Rightarrow$ put $x+1=t$

$dx=dt$

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

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

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

$=\frac{1}{2}\cdot \frac{1}{2t}\sqrt{{\left(\frac{1}{t}\right)}^2}+1+\frac{1}{2}ln|\frac{1}{t}+\sqrt{\frac{1}{t^2}+1}|-\frac{t}{2}-\frac{1}{2}ln|t|+c$

$=\frac{1}{4t}\sqrt{{\left(\frac{1}{t}\right)}^2+1}+ln\left|\sqrt{\frac{1}{t}+\sqrt{\frac{1}{t^2}+1}}\right|-\frac{t}{2}-ln|\sqrt{t}|+c$

$=\frac{1}{2t}({\left(\frac{1}{t}\right)}^2+1)+ln\left|\sqrt{\frac{1}{t}+\sqrt{\frac{1}{t^2}+1}}\right|-\frac{t}{2}-ln|\sqrt{t}|+c$

$=\frac{1}{2(x+1)}[\frac{1}{(x+1{)}^2}+1]+ln|\sqrt{\frac{1}{x+1}+\sqrt{\frac{1}{(x+1{)}^2}}}+1|-\frac{x+1}{2}-ln\left|\sqrt{x+1}\right|+c$