If the solve the problem


If $\int \sin ^{-1}\left(\sqrt{\frac{x}{1+x}}\right) d x=A(x) \tan ^{-1}(\sqrt{x})+B(x)+C$

where $\mathrm{C}$ is a constant of integration, then the ordered pair $(\mathrm{A}(\mathrm{x}), \mathrm{B}(\mathrm{x}))$ can be :

  1. $(x-1, \sqrt{x})$

  2. $(x+1, \sqrt{x})$

  3. $(x+1,-\sqrt{x})$

  4. $(x-1,-\sqrt{x})$

Correct Option: , 3


Put $\quad x=\tan ^{2} \theta \Rightarrow d x=2 \tan \theta \sec ^{2} \theta d \theta$

$\int \theta \cdot\left(2 \tan \theta \cdot \sec ^{2} \theta\right) \mathrm{d} \theta$

$\downarrow \quad \downarrow$

I $\quad$ II $\quad$ (By parts)

$=\theta \cdot \tan ^{2} \theta-\int \tan ^{2} \theta d \theta$

$=\theta \cdot \tan ^{2} \theta-\int\left(\sec ^{2} \theta-1\right) \mathrm{d} \theta$

$=\theta\left(1+\tan ^{2} \theta\right)-\tan \theta+C$

$=\tan ^{-1}(\sqrt{x})(1+x)-\sqrt{x}+C$

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