Chapter 7: Integrals

Ans: Let $I = \int {\sqrt {{x^2} - 8x + 7} } dx$

$I = \int {\sqrt {\left( {{x^2} - 8x + 16} \right) - 9} dx} $

$I = \int {\sqrt {{{\left( {x - 4} \right)}^2} - {{\left( 3 \right)}^2}} dx} $

It is known that,

$\sqrt {{x^2} - {a^2}} dx = \dfrac{x}{2}\sqrt {{x^2} - {a^2}}  - \dfrac{{{a^2}}}{2}\log \left| {x + \sqrt {{x^2} - {a^2}} } \right| + C$ 

For $a$ substitute $3$ and for $x$ put $x - 4$ and solve the integral.

$\therefore I = \dfrac{{\left( {x - 4} \right)}}{2}\sqrt {{x^2} - 8x + 7}  - \dfrac{9}{2}\log \left| {\left( {x - 4} \right) + \sqrt {{x^2} - 8x + 7} } \right| + C$

Hence, the correct answer is D.