MCQ
$\int \sqrt{x^{2}-8 x+7} d x$ बराबर है
  • A
    $\frac{1}{2}(x-4) \sqrt{x^{2}-8 x+7}$ $+9 \log \left|x-4+\sqrt{x^{2}-8 x+7}\right|$ + C
     
  • B
    $\frac{1}{2}(x-4) \sqrt{x^{2}-8 x+7}$ $-\frac{9}{2} \log \left|x-4+\sqrt{x^{2}-8 x+7}\right|$ + C
  • C
    $\frac{1}{2}(x-4) \sqrt{x^{2}-8 x+7}$ $-3 \sqrt{2} \log \left|x-4+\sqrt{x^{2}-8 x+7}\right|$ + C
  • D
    $\frac{1}{2}(x+4) \sqrt{x^{2}-8 x+7}$ $+9 \log \left|x+4+\sqrt{x^{2}-8 x+7}\right|+C$

Answer

माना $I=\int \sqrt{x^{2}-8 x+7} d x \Rightarrow I$ $=\int \sqrt{x^{2}-8 x+7+(4)^{2}-(4)^{2}} d x$
$=\int \sqrt{(x-4)^{2}+7-16} d x$ $=\int \sqrt{(x-4)^{2}-(3)^{2}} d x$ ($\because$ $\int \sqrt{x^{2}-a^{2}} d x$ $=\frac{x}{2} \sqrt{x^{2}-a^{2}}$ $-\frac{a^{2}}{2} \log \left|x+\sqrt{x^{2}-a^{2}}\right|$
$\Rightarrow$ $I=\frac{x-4}{2} \sqrt{x^{2}-8 x+7}$ $-\frac{9}{2} \log \left|x-4+\sqrt{x^{2}-8 x+7}\right|$ + C

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