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Question:

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

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

 

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

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

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

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


Correct Option: 1

Solution:

$I=\int \sin ^{-1}\left(\frac{\sqrt{x}}{\sqrt{1+x}}\right) d x=\int \tan ^{-1} \sqrt{x} \cdot 1 d x$

$=x \tan ^{-1} \sqrt{x}-\int \frac{1}{1+x} \cdot \frac{1}{2 \sqrt{x}} \cdot x d x+C$

$=x \tan ^{-1} \sqrt{x}-\frac{1}{2} \int \frac{t \cdot 2 t d t}{1+t^{2}}+C$ (Put $x=t^{2} \Rightarrow d x=2 t d t$ )

$=x \tan ^{-1} \sqrt{x}-\int \frac{t^{2}}{1+t^{2}} d t+C$

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

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

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

$\Rightarrow A(x)=x+1 \Rightarrow B(x)=-\sqrt{x}$

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