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Integrals — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsIntegrals1 Mark
MCQ
Choose the correct answer in Exercise.$\int\sqrt{\text{x}^2-8\text{x}+7}\text{dx}$ is equal to
  • A
    $\frac{1}{2}(\text{x}-4)\sqrt{\text{x}^2-8\text{x}+7}+9\text{log}\Bigg|\text{x}-4+\sqrt{\text{x}^2-8\text{x}+7}\Bigg|+\text{C}$
  • B
    $\frac{1}{2}(\text{x}+4)\sqrt{\text{x}^2-8\text{x}+7}+9\text{log}\Bigg|\text{x}+4+\sqrt{\text{x}^2-8\text{x}+7}\Bigg|+\text{C}$
  • C
    $\frac{1}{2}(\text{x}-4)\sqrt{\text{x}^2-8\text{x}+7}-3\sqrt2\text{log}\Bigg|\text{x}-4+\sqrt{\text{x}^2-8\text{x}+7}\Bigg|+\text{C}$
  • ✓
    $\frac{1}{2}(\text{x}-4)\sqrt{\text{x}^2-8\text{x}+7}-\frac{9}{2}\text{log}\Bigg|\text{x}-4+\sqrt{\text{x}^2-8\text{x}+7}\Bigg|+\text{C}$
✓

Answer

Correct option: D.
$\frac{1}{2}(\text{x}-4)\sqrt{\text{x}^2-8\text{x}+7}-\frac{9}{2}\text{log}\Bigg|\text{x}-4+\sqrt{\text{x}^2-8\text{x}+7}\Bigg|+\text{C}$
$=\Bigg(\frac{\text{x}-4}{2}\Bigg)\sqrt{(\text{x}-4)^2-3^2}-\frac{3^2}{2}\text{log}\Big|\text{x}-4+\sqrt{(\text{x}-4)^2-3^2}\Big|+\text{C}$
$\Bigg[\therefore\ \int\sqrt{\text{x}^2-\text{a}^2}\text{dx}=\frac{\text{x}}{2}\sqrt{\text{x}^2-\text{a}^2}-\frac{\text{a}^2}{2}\text{log}\Big|\text{x}+\sqrt{\text{x}^2-\text{a}^2}\Big|\Bigg]$
$=\Bigg(\frac{\text{x}-4}{2}\Bigg)\sqrt{\text{x}^2-8\text{x}+7}-\frac{9}{2}\text{log}\Big|\text{x}-4+\sqrt{\text{x}^2-8​​\text{x}+7}\Big|+\text{C}$

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