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STD 12 - 7.1 inderfinite integral — Mathematics STD 12 Science — Question

CBSE BoardEnglish MediumSTD 12 ScienceMathematicsSTD 12 - 7.1 inderfinite integral1 Mark
MCQ
$\int_{}^{} {\sqrt {{x^2} - 8x + 7} } \;dx = $
  • A
    $\frac{1}{2}(x - 4)\sqrt {{x^2} - 8x + 7} + 9\log [x - 4 + \sqrt {{x^2} - 8x + 7} ] + c$
  • B
    $\frac{1}{2}(x - 4)\sqrt {{x^2} - 8x + 7} - 3\sqrt 2 \log [x - 4 + \sqrt {{x^2} - 8x + 7} ] + c$
  • ✓
    $\frac{1}{2}(x - 4)\sqrt {{x^2} - 8x + 7} - \frac{9}{2}\log [x - 4 + \sqrt {{x^2} - 8x + 7} ] + c$
  • D
    None of these
✓

Answer

Correct option: C.
$\frac{1}{2}(x - 4)\sqrt {{x^2} - 8x + 7} - \frac{9}{2}\log [x - 4 + \sqrt {{x^2} - 8x + 7} ] + c$
c
(c)$\int_{}^{} {\sqrt {{x^2} - 8x + 7} \,dx = \int_{}^{} {\sqrt {{{(x - 4)}^2} - {{(3)}^2}} \,dx} } $
Now apply formula of $\int_{}^{} {\sqrt {{x^2} - {a^2}} \,dx.} $

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