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Integrals — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsIntegrals1 Mark
MCQ
Choose the correct option from given four options : $\int\tan^{-1}\sqrt{\text{x}}\text{ dx}$ is equal to :
  • ✓
    $(\text{x}+1)\tan^{-1}\sqrt{\text{x}}-\sqrt{\text{x}}+\text{C}$
  • B
    $\text{x}\tan^{-1}\sqrt{\text{x}}-\sqrt{\text{x}}+\text{C}$
  • C
    $\sqrt{\text{x}}-\text{x}\tan^{-1}\sqrt{\text{x}}+\text{C}$
  • D
    $\sqrt{\text{x}}-(\text{x}+1)\tan^{-1}\sqrt{\text{x}}+\text{C}$
✓

Answer

Correct option: A.
$(\text{x}+1)\tan^{-1}\sqrt{\text{x}}-\sqrt{\text{x}}+\text{C}$
Let $\text{I}=\int1\cdot\tan^{-1}\sqrt{\text{x}}\text{ dx}$
$=\text{x}\tan^{-1}\sqrt{\text{x}}-\int\text{x}\cdot\frac{1}{1+\text{x}}\frac{1}{2\sqrt{\text{x}}}\text{dx}$
$=\text{x}\tan^{-1}\sqrt{\text{x}}-\frac{1}{2}\int\sqrt{\text{x}}\cdot\frac{1}{1+\text{x}}\text{dx}$
Put $\text{x}=\text{t}^2$
$\Rightarrow\text{dx}=2\text{t dt}$
$\therefore$ we get ;
$=\text{x}\tan^{-1}\sqrt{\text{x}}-\int\frac{\text{t}^2}{1+\text{t}^2}\text{dt}$
$=\text{x}\tan^{-1}\sqrt{\text{x}}-\int\Big(1-\frac{1}{1+\text{t}^2}\Big)\text{dt}$
$=\text{x}\tan^{-1}\sqrt{\text{x}}-\text{t}+\tan^{-1}\text{t}+\text{C}$
$=\text{x}\tan^{-1}\sqrt{\text{x}}-\sqrt{\text{x}}+\tan^{-1}\text{t}+\text{C}$
$=\text{x}\tan^{-1}\sqrt{\text{x}}-\sqrt{\text{x}}+\tan^{-1}\sqrt{\text{x}}+\text{C}$
$=(\text{x}+1)\tan^{-1}\sqrt{\text{x}}-\sqrt{\text{x}}+\text{C}$

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