Integrals — Class 12 Maths Solution

: Let $I = \int {\cfrac{{dx}}{{\sqrt {{{\left( {2 - x} \right)}^2} + 1} }}}$

Put $\left( {2 - x} \right) = t$ $\Rightarrow$ $- dx = dt$ $\Rightarrow$ $dx = - dt$

$\therefore$ $I = - \int {\cfrac{{dt}}{{\sqrt {{t^2} + 1} }}} = - \log \left| {t + \sqrt {{t^2} + 1} } \right| + C$

$= - \log \left| {\left( {2 - x} \right) + \sqrt {{{\left( {2 - x} \right)}^2} + 1} } \right| + C$

$= \log \left| {\cfrac{1}{{\left( {2 - x} \right) + \sqrt {{x^2} - 4x + 5} }}} \right| + C$