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INTEGRALS — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsINTEGRALS1 Mark
Question
Choose the correct answer in Exercise.
$\int\sqrt{1+\text{x}^2}\text{dx}$ is equal to
  1. $\frac{\text{x}}{2}\sqrt{1+\text{x}^2}+\frac{1}{2}\text{log}\Bigg|\Big(\text{x}+\sqrt{1+\text{x}^2}\Big)\Bigg|+\text{C}$
  2. $\frac{2}{3}(1+\text{x}^2)^{\frac{3}{2}}+\text{C}$
  3. $\frac{2}{3}\text{x}(1+\text{x}^2)^{\frac{3}{2}}+\text{C}$
  4. $\frac{\text{x}^2}{2}\sqrt{1+\text{x}^2}+\frac{1}{2}\text{x}^2\text{log}\Bigg|\text{x}+\sqrt{1+\text{x}^2}\Bigg|+\text{C}$
✓

Answer

  1. $\frac{\text{x}}{2}\sqrt{1+\text{x}^2}+\frac{1}{2}\text{log}\Bigg|\Big(\text{x}+\sqrt{1+\text{x}^2}\Big)\Bigg|+\text{C}$

$\int\sqrt{1+\text{x}^2}\text{dx}=\int\sqrt{\text{x}^2+1^2}\text{dx}=\frac{\text{x}}{2}\sqrt{\text{x}^2+1^2}+\frac{1^2}{2}\text{log}\Big|\text{x}+\sqrt{\text{x}^2+1^2}\Big|+\text{C}$

$=\frac{\text{x}}{2}\sqrt{\text{x}^2+1}+\frac{1}{2}\text{log}\Big|\text{x}+\sqrt{\text{x}^2+1}\Big|+\text{C}$

$\Bigg[\therefore\ \int\sqrt{\text{x}^2+\text{a}^2}\text{dx}=\frac{\text{x}}{2}\sqrt{\text{x}^2+\text{a}^2}+\frac{\text{a}^2}{2}\text{log}\Big|\text{x}+\sqrt{\text{x}^2+\text{a}^2}\Big|\Bigg]+\text{c}$

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