Question

If $y=\int \frac{dx}{(1+x^2{)}^{3/2}}$ and $y=0$ when $x=0$, then value of $y$ when $x=1$ , is

• A. None of these
• B. $\sqrt{2}$
• C. $2\sqrt{2}$
• D. $\frac{1}{\sqrt{2}}$

Answer 1

The correct option is D $\frac{1}{\sqrt{2}}$

$y=\int \frac{dx}{{\left(1+x^2\right)}^{3/2}}\Rightarrow y=\int \frac{dx}{x^3{\left(1+\frac{1}{x^2}\right)}^{3/2}}$

$Let1+\frac{1}{x^2}=t-2\frac{1}{x^3}dx=dt\therefore dx=\frac{x^3}{-2}dt$

Therefore,

$\Rightarrow y=\int \frac{dx}{-2\left(t^{3/2}\right)}\Rightarrow y=\frac{1}{-2}\times \frac{t^{-3/2+1}}{\left(\frac{-3}{2}+1\right)}+c\Rightarrow y=t^{-1/2}+c\Rightarrow y=\frac{1}{{\left(1+\frac{1}{x^2}\right)}^{1/2}}+c\to \left(i\right)$

$Putx=0andy=0c=0andputx=1inequation\left(i\right)y=\frac{1}{\sqrt{2}}$

Hence, this is the answer.