Question

If $y=\int \frac{dx}{(1+x^2{)}^{\frac{1}{2}}}$ and $y=0$ when $x=0$, then value of $y$ when$x=1$, is:

• A. $\ln \left(2\right)$
• B. $\ln \left(\sqrt{2}\right)$
• C. $\ln (1+\sqrt{2})$
• D. $\frac{1}{\sqrt{2}}$

Answer 1

Answer: (C)

$\ln (1+\sqrt{2})$

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

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

$⟹y=\int \frac{dt}{t}=\ln t+C=\ln (x+\sqrt{x^2+1})+C$

when $x=0$ ,$y=0⟹\ln \left(1\right)+C=0⟹C=0$

$y=\ln (x+\sqrt{x^2+1})$

when $x=1,y=\ln (1+\sqrt{(1{)}^2+1})=\ln (1+\sqrt{2})$