Question

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

• A. $\log |x-2+\sqrt{{(x-1)}^2+2}|$
• B. $\log |x-2+\sqrt{{(x-2)}^2+1}|$
• C. $\log |x-1+\sqrt{{(x-1)}^2+2}|$
• D. $\log |x-2+\sqrt{{(x-1)}^2+1}|$

Answer 1

The correct option is C $\log |x-1+\sqrt{{(x-1)}^2+2}|$

We know that $\int \frac{1}{\sqrt{x^2+a^2}}dx$ = $\log |x+\sqrt{x^2+a^2}|$
Here , instead of x we have x - 1 and the value of a will be =
$\sqrt{2}$
So $\int \frac{1}{\sqrt{(x-1{)}^2+(\sqrt{2}{)}^2}}dx$ would be equal to $log|(x-1)+\sqrt{(x-1{)}^2+2}|+C$

Hence, $q=\sqrt{(x-1{)}^2+2}$